Topology and Functional Analysis

MAS8220: Topology and Functional Analysis

Dr. Zinaida Lykova

School of Mathematics and Statistics Newcastle University Room 3.13 in Herschel Building * This subject constitutes a synthesis of some of the main trends in analysis over the past century. One studies functions not individually, but as a collection which admits natural operations of addition and multiplication and has geometric structure. An algebra is a vector space with an associative multiplication. There is an abundance of natural examples, many of them having the structure of a Banach space. Examples are the spaces of n n matrices and the continuous functions on × the interval [0, 1], with suitable norms. Putting together algebras and norms one is led to the idea of a Banach algebra. A rich and elegant theory of such objects was developed over the second half of the twentieth century. Several members of staﬀ have research interests close to this area.

You can consult the following books.

1. F. F. Bonsall, J. Duncan, “Complete Normed Algebras”. Springer-Verlag, New York, 1973.

2. G. K. Pedersen, “Analysis now”, Springer-Verlag, 1989.

3. N. J. Young, “An Introduction to Hilbert Space”, Cambridge University Press, 1988.

4. W. Rudin, “Functional analysis”, 2nd Edition, McGraw-Hill, 1991.

5. S. Willard, “General topology”, Dover Publications, 2004.

2 Teaching Information - MAS8220 - 2015

Lectures:

Monday 15.00 – 17.00 HERB.4 TR2 Thursday 15.00 – 16.00 HERB.4 TR2

Tutorials:

Thursday 16.00 – 17.00 HERB.4 TR2

Oﬃce hours:

Monday 17.00 – 18.00 HERB.L3 Room 3.13. Tuesday 14.00 – 15.00 HERB.L3 Room 3.13.

Revision Lectures:

Monday 11th May 2015, 15.00-17.00, HERB.L4 TR2. Thursday 14th May 2015, 15.00-16.00, HERB.L4 TR2.

Thursday 14th May 2015, 16.00-17.00, HERB.L4 TR2 (Drop-in)

Assessment:

1. Assessment of course work (10%).

Assignments:

Assignments are to be handed in by 4pm on Tuesdays: 10 February (no. 1), 24 February (no. 2, 3), 10 March (no. 4, 5), 21 April (no. 6, 7), 5 May 2015 (no. 8).

2. 2 hours 15 min examination at the end of Semester (90%).

3 Part I. Banach algebras and continuous homomorphisms

This part concerns the theory of Banach algebras and continuous homomorphisms between Banach algebras. We shall deﬁne these terms and study their properties.

Contents of Part I

Pages

1. Ring and algebras 5

2. Normed algebras 7

3. Spectra and regular elements 11

4. Analytic vector functions and the resolvent 17

4.1. Cauchy’s integral formula 19

5. The Gelfand-Mazur Theorem 22

6. Convolution algebras 24

7. Homomorphisms and ideals 29

8. The algebra C(K) 36

9. Characters of C(K) 39

4 1 Rings and algebras

Examples of rings: Z, C, C[z], Mn(C). Deﬁnition 1.1. A ring is a triple (R, +, ) where R is a set and +, are binary operations on R such that

(R, +) is an abelian group; • multiplication is an associative operation: a, b, c R, • ∀ ∈ a (b c)=(a b) c; multiplication is distributive over +: a, b, c R, • ∀ ∈ a (b + c)=(a b)+(a c), (b + c) a =(b a)+(c a). Notes 1.2. 1. We call +, addition and multiplication respectively. We often write ab for a b.

2. Every ﬁeld is a ring: thus R, C, Z5 with the usual addition and multiplication are rings.

3. The set 2Z of even integers, with the usual addition and multiplication, is a ring. It does not have an identity element for multiplication (a ”one”).

4. In the ring M (R) of 2 2 matrices over R, with the usual addition and 2 × multiplication, a b is typically not equal to b a. M2(R) is a non-commutative ring.

Definition 1.3. We say that (R, +, ) is a ring with identity if there is an element e R such that x e = e x = xfor allx R. ∈ ∈ Definition 1.4. (R, +, ) is said to be a commutative ring if a b = b a for all a, b R. ∈ Examples 1.5. 1. C[x, y], the set of polynomials in the two variables x, y with complex coefficients, is a ring with respect to the usual addition and multiplication of polynomials. It is commutative and has an identity, which is the constant polynomial 1.

2. Z[x], the set of polynomials in the variable x with integer coeﬃcients, is a commutative ring with identity.

5 3. Z6, the set of residue classes of integers mod 6, with the usual addition and multiplication, is a commutative ring with identity. It is not a ﬁeld since [2] has no multiplicative inverse.

Recall that a ﬁeld is a commutative ring with identity in which every non- zero element has a multiplicative inverse.

4. Tn, the set of upper triangular complex n n matrices with the usual matrix addition and multiplication is a non-commutative× ring with identity (where n> 1).

In checking a statement like (4) we do not check every ring axiom. We observe that Tn is a subset of the standard ring Mn(C), and use the following observation. Lemma 1.6. If R is a ring and S is a subset of R such that (i) 0 S, where 0 is the zero of R; ∈ (ii) for each x S, the additive inverse of x is also in S. ∈ (iii) S is closed under addition and multiplication, then S is a ring with respect to the same algebraic operations as R. Among the above examples of rings, some are also vector spaces, some not.

Deﬁnition 1.7. Let k be a ﬁeld. An algebra over k is a vector space A over k equipped with a binary operation such that (i) (A, +, ) is a ring; (ii) λ(x y)=(λx) y = x (λy) for all λ k and x, y A. ∈ ∈ Examples 1.8. 1. Mn(C) is an algebra over C.

2. R[x] is an algebra over R.

3. Z is a ring, but is not an algebra over R or C : there is no natural way to deﬁne λn Z for general λ R and n Z. ∈ ∈ ∈

4. Z[x] is a ring, but not an algebra over R or C.

6 2 Normed algebras

These are algebras over R or C which are also normed space.

Examples: C[0, 1], Mn(C).

Recall that a normed linear subspace is a vector space V over R or C together with a mapping : V R+ such that, for any x, y V and λ R or C, the → ∈ ∈ following conditions are satisﬁed 1. x = 0 if and only if x = 0; 2. λx = λ x ; | | 3. x + y x + y . ≤

Deﬁnition 2.1. A normed algebra is a normed linear space (A, ) equipped with a binary operation (x, y) xy which makes A into an algebra and satisﬁes → xy x y for all x, y A. (2.1) ≤ ∈ Property (2.1) is called the ”submultiplicativity of the norm”.

Examples 2.2. 1. (C, . ) is a normed algebra over C. | | 2. (C[0, 1], . ) is a normed algebra over C, where C[0, 1] is the set of contin- ∞ uous C valued functions on [0, 1], addition and multiplication are the usual ”pointwise”− operations,

(x + y)(t)= x(t)+ y(t), (xy)(t)= x(t)y(t), for all x, y C[0, 1], t [0, 1], ∈ ∈ and the norm x = sup x(t) . ∞ 0 t 1 | | ≤ ≤ Check ”submultiplicative property”. Consider x, y C[0, 1]. ∈ xy = sup (xy)(t) ∞ 0 t 1 | | ≤ ≤ = sup x(t) y(t) 0 t 1 | || | ≤ ≤ sup x(t) sup y(t) ≤ 0 t 1 | | 0 t 1 | | ≤ ≤ ≤ ≤ = x y . ∞ ∞

7 3. (C[z], . ) is a normed algebra over C, where C[z] denotes the set of poly- ∞ nomials over C and addition, multiplication and . are as in (2). ∞ 4. (M (C), . ), the algebra of n n matrices over C is a normed algebra over n × C, where . is the ”operator norm”:

T = sup T x Cn x Cn 1 ≤ n Here . n is the usual Euclidean norm on C C 1 n 2 2 x n = x . C | j| ( j=1 ) X Check submultiplictivity. Consider S, T M (C). ∈ n

ST = sup ST x Cn x Cn 1 ≤

sup S T x Cn ≤ x Cn 1 ≤

= S sup T x Cn x Cn 1 ≤ = S T .

Recall that a normed linear space is said to be complete if every Cauchy sequence converges in the space.

A Cauchy sequence in (X, . ) is a sequence (x ) in X such that, for every n ǫ> 0 there exists N N such that ∈ x x < ǫ whenever m, n N. n − m ≥ A Banach space is a complete normed linear space.

Deﬁnition 2.3. A Banach algebra is a normed algebra over R or C that is complete as a normed linear space.

We shall only study complex Banach algebras, i.e. Banach algebras over C.

Examples 2.4. 1. (C[0, 1], . ) is a Banach space and a normed algebra, ∞ hence it is a Banach algebra. (cf. Example 2.2(2).)

8 2. (C[z], . ) is a subalgebra of C[0, 1]. It is a normed algebra, but is incom- ∞ plete, so not a Banach algebra.

Claim: (C[z], . ) is incomplete. Consider the sequence (xn) in C[z], ∞ where z2 zn x (z)=1+ z + + ... + . n 2! n! If m > n then zn+1 zm xm xn = sup + ... + − ∞ 0 z 1 (n + 1)! m! ≤ ≤

1 1 + ... + . ≤ (n + 1)! m!

It is easy to see that (xn) is Cauchy for . , but ∞ x exp as n n → →∞

in (C[0, 1], . ). Therefore xn cannot tend to a polynomial, i.e. (xn) is a ∞ non-convergent Cauchy sequence in (C[z], . ). Hence (C[z], . ) is in- ∞ ∞ complete.

3. Mn(C) is a ﬁnite-dimensional normed algebra. It is well-known that every ﬁnite-dimensional normed linear space over C or R is complete. Thus Mn(C) is complete, so is a Banach algebra.

4. ”Zero multiplication algebras”. Any Banach space E becomes a Banach algebra if we deﬁne xy to be 0 for all x, y E. ∈

Multiplication in a normed algebra A is a mapping from A A to A. We can make A A into a normed linear space by deﬁning, for x ,y , x ,y× A and λ C, × 1 1 2 2 ∈ ∈

(x1,y1)+(x2,y2)=(x1 + x2,y1 + y2)

λ(x1,y1)=(λx1,λy1)

(x1,y1) A A = max x1 A, y1 A . × { }

9 Similarly C A is a normed linear space with respect to × (λ, x) C A = max λ , x A × {| | } for λ C, x A. ∈ ∈ Theorem 2.5. In any normed algebra A, multiplication is a continuous mapping from A A to A; scalar multiplication is a continuous mapping from C A to A. × × Proof. Consider (x ,y ) A A. Let ǫ> 0. Observe that, for x, y A, 0 0 ∈ × ∈ xy x y = x(y y )+(x x )y − 0 0 − 0 − 0 0 x y y + x x y . ≤ − 0 − 0 0 Let ǫ> 0, choose ǫ δ = min 1, 1 2(1 + y ) 0 ǫ δ = min 1, 2 2(1 + x ) 0 δ = min δ , δ . { 1 2}

Suppose (x, y) (x0,y0) A A < δ, that is, max x x0 A, y y0 A < δ. × Then − { − − } x x < δ 1, − 0 1 ≤ so that x < 1+ x . 0 Furthermore ǫ x x0 < , − 2(1 + y0 ) ǫ y y < . − 0 2(1 + x ) 0 Hence

xy x0y0 x y y0 + x x0 y0 − ≤ − ǫ − ǫ < (1 + x0 ) +(1+ y0 ) 2(1 + x0 ) 2(1 + y0 ) ǫ ǫ + ≤ 2 2 = ǫ.

Hence multiplication is continuous at (x0,y0).

Scalar multiplication - exercise.

10 3 Spectra and regular elements

The spectrum of an n n matrix T is the set of its eigenvalues, that is, the set of all λ C such that λI× T is singular. The notion extends fruitfully to Banach algebras.∈ −

In this section we consider a Banach algebra A with identity element e. We assume e =0. (Think of C[0, 1] and M (C).) n Deﬁnition 3.1. An element x A is said to be a regular element (or invertible ∈ element) if there exists y A such that xy = e and yx = e. ∈ An element x A is singular if it is not regular. Any such y is called an inverse of x. ∈

Note that x can have at most one inverse, for if y1 and y2 are inverses of x then

y1 = y1e = y1(xy2)=(y1x)y2 = ey2 = y2.

1 We may speak of ”the” inverse of a regular element x; we denote it by x− . Examples 3.2. 1. The element

1 2 T = 1 3 4

is regular in M2(C); the element

1 2 T = 2 3 6

is singular in M2(C).

2. In C[0, 1],

f : [0, 1] C : t 1+ t is regular, with the inverse t 1 . → → → 1+t

g : [0, 1] C : t t is singular. → →

f is regular in C[0, 1] if and only if 0 / Rangef. ∈

11 Engineers call the following statement the ”small gain theorem”.

Lemma 3.3. Let A be a Banach algebra with the identity e. If x A and e x < 1 then x is a regular element of A and ∈ −

1 1 x− e e x . − ≤ 1 e x − − − Proof. Let u = e x, so that − x = e u and u < 1. − We shall show that 1 1 2 x− =(e u)− = e + u + u + ... (3.1) − First show that the RHS of (3.1) is meaningful, that is, deﬁnes an element of A.

For n N, let ∈ 2 n yn = e + u + u + ... + u .

Claim: (yn) is a Cauchy sequence in A.

For m > n,

y y = un+1 + ... + um, m − n and so

y y u n+1 + ... + u m m − n≤ u n+1 (since u < 1). ≤ 1 u −

Thus ym yn 0 as m, n . Therefore (yn) converges to a limit y A. Furthermore − → → ∞ ∈

xy =(e u)y = y uy n − n n − n = e + u + ... + un (u + u2 + ... + un+1) − = e un+1 − ∴ xy e = un+1 u n+1 n − ≤ and so, since u < 1, xy e as n . n → →∞

12 Since y y and multiplication is continuous, n →

xy = x lim yn = lim xyn = e. n n

1 Similarly yx = e. Thus x is a regular element. Furthermore x− = y, and

1 x− e = y e = lim(yn e) − − n − = lim(u + u2 + ... + un). n

1 2 n ∴ x− e = lim(u + u + ... + u ) − n = lim u + u2 + ... + un n lim u + u 2 + ... + u n ≤ ∞ = u j j=1 X u = since u < 1 1 u − e x = − . 1 e x − −

Deﬁnition 3.4. Let X be a normed linear space. For x X and ǫ> 0 the open ∈ ball of centre x, radius ǫ is the set

y X : x y < ǫ . { ∈ − } and is denoted by Bǫ(x). A subset G of X is said to be open if, for every x G, there exists ǫ> 0 such that B (x) G. A closed set is a set whose complement∈ is open. ǫ ⊂ Theorem 3.5. Let A be a Banach algebra with the identity e and let G(A) be the set of regular elements in A. Then 1. G(A) is a group under multiplication;

2. G(A) is an open set in A;

3. the inversion map 1 x x− → is a continuous map from G(A) to G(A).

13 1 1 1 1 Proof. 1. If x, y G(A) then x− ,y− exist and y− x− satisﬁes ∈ 1 1 1 1 (xy)(y− x− )= e =(y− x− )(xy),

1 1 1 so that xy G(A) and (xy)− = y− x− . ∈

The element e is clearly a multiplicative identity for G(A).

1 1 If x G(A) then x− has inverse x in A, so x− G(A). ∈ ∈

Multiplication is associative. Hence G(A) is a group with respect to multiplication.

2. Consider any x G(A). To prove that G(A) is open in A, we must ﬁnd ǫ> 0 ∈ 1 1 such that Bǫ(x) G(A). Let ǫ = x− − > 0. Consider any y Bǫ(x), then 1 ⊂1 ∈ x y < x− − . Thus we have − 1 1 1 1 1 1 e x− y = x− (x y) x− x y < x− x− − =1. − − ≤ −

1 By Lemma 3.3, we have x− y G(A). ∈

1 ∴ x(x− y) G(A). ∈

∴ y G(A). ∈

∴ B (x) G(A). ǫ ⊂

Hence G(A) is open.

3. Inversion is continuous Let 1 f : G(A) G(A): x x− . → → We want to show that if x x in G(A) then f(x ) f(x). n → n → (i). First consider a sequence (xn) in G(A) such that xn e. By Lemma 3.3, if e x < 1, → − n

1 f(x ) e = x− e n − n − 14 e x − n ≤ 1 e x − − n 0 as x e. → n → 1 Now suppose xn x in G(A). Then x− xn e, by continuity of multiplication. → → By Part (i), 1 1 (x− x )− e n → 1 as n , that is, x− x e as n , →∞ n → →∞ 1 1 1 1 1 ∴ x− = x− xx− ex− = x− , n n → 1 again by continuity of multiplication. Hence f : x x− is continuous. →

Deﬁnition 3.6. The spectrum of an element x A is the set ∈ λ C : λe x is singular in A . { ∈ − } It is denoted by σ(x). The resolvent set of x, denoted by ρ(x), is the complement of σ(x): ρ(x)= λ C : λe x is regular in A . { ∈ − } Examples 3.7. (i) For T M (C), σ(T ) is the set of eigenvalues of T . ∈ n (ii) For f C[0, 1], σ(f) = Range f. ∈ Theorem 3.8. Let A be a Banach algebra with the identity e. For any x A, ∈ 1. λ σ(x) λ x ; ∈ ⇒| |≤ 2. ρ(x) is open in C;

3. σ(x) is a compact subset C.

Proof. 1. We show that λ > x λ / σ(x), that is, λ ρ(x). | | ⇒ ∈ ∈

Suppose λ C and λ > x . Then λ = 0 and x/λ < 1, and so, by Lemma 3.3,∈e x/λ is| regular| . −

15 ∴ λ(e x/λ) is regular. −

∴ λe x is regular. −

∴ λ ρ(x). ∈ Hence λ σ(x) λ x . Thus σ(x) is a bounded set in C. ∈ ⇒| |≤ 2. Let x A and deﬁne ∈ F : C A : λ λe x. → → −

For λ, C, we have ∈ F (λ) F () = λe e = λ e , − − | − | and hence F is continuous. By deﬁnition,

1 ρ(x)= λ C : F (λ) is regular in A = F − (G(A)). { ∈ } Since G(A) is open and F is continuous, ρ(x) is open.

3. Since ρ(x) is open and σ(x)= C ρ(x), σ(x) is closed. The spectrum σ(x) is also bounded. Thus, by the Bolzano-Weierstrass\ theorem, σ(x) is compact.

16 4 Analytic vector functions and the resolvent

Recall that, for an open set Ω C, a function f : Ω C is analytic if f is complex diﬀerentiable at every point of⊂ Ω, that is, if the limit→ f(z + h) f(z) lim − exists for each z Ω. h 0,h C → ∈ h ∈ In other words, if z Ω there exists f ′(z) C such that ∀ ∈ ∈ ǫ> 0 δ > 0 such that ∀ ∃ f(z + h) f(z) − f ′(z) < ǫ h −

whenever h < δ and z + h Ω. | | ∈ We can adapt there notions with minimal modifications to vector-valued functions. Let X be a Banach space and consider a function f : Ω X. → For example, we might think of T M (C) and ∈ n 1 f : ρ(T ) M (C): λ (λI T )− . → n → − Definition 4.1. We say that f : Ω X is differentiable at z Ω if → ∈ f(z + h) f(z) lim − exists, h 0,h C → ∈ h or in other words, if there exists f ′(z) X such that for all ǫ > 0 there exists ∈ δ > 0 such that f(z + h) f(z) − f ′(z) < ǫ h − X whenever h < δ and z + h Ω. | | ∈ We say that f is analytic or holomorphic in Ω if f is differentiable at every point of Ω. Theorem 4.2. Let A be a Banach algebra with identity e and let x A. For λ ρ(x), let ∈ ∈ 1 x(λ)=(λe x)− . − Then the function x( ): ρ(x) A → is analytic in ρ(x) and, for all λ, ρ(x), ∈ x() x(λ)=(λ )x(λ)x(). (4.1) − −

17 Proof. By Theorem 3.5(3) and Theorem 3.8(2), x( ): ρ(x) A is a continuous function, being the composition of two known continuous functions→

1 λ λe x (λe x)− . → − → − ρ(x) G(A) G(A). → → For λ, ρ(x), ∈ 1 1 x(λ)− x()− = λe x (e x) − − − − =(λ )e. − Therefore

1 1 x(λ)(x(λ)− x()− )x()= x(λ)(λ )ex() − −

∴ x() x(λ)=(λ )x(λ)x(). − − Thus (4.1) holds. It is called the resolvent identity. Next we show that x( ) is diﬀerentiable at λ ρ(x). Put = λ + h ρ(x). By (4.1), ∈ ∈ x(λ + h) x(λ)= hx(λ)x(λ + h). − − x(λ + h) x(λ) ∴ − ( x(λ)2)= x(λ)(x(λ) x(λ + h)) h − − −

x(λ + h) x(λ) ∴ − ( x(λ)2) x(λ) x(λ) x(λ + h) . h − − ≤ A − A A

RHS 0 as h 0. Hence → → x(λ + h) x(λ) lim − = x(λ)2. h 0 h → − 2 Thus x′(λ) exists and equals x(λ) for any λ ρ(x). Thus x( ) is analytic on ρ(x). − ∈

Deﬁnition 4.3. The function x( ): ρ(x) A is called the resolvent of x. →

18 4.1 Cauchy’s integral formula If X is a Banach space and f is an analytic X-valued function on an open set U C then we may deﬁne contour integrals ⊂ f(z)dz, Zγ for a contour γ :[a, b] U, just as for scalar-valued functions. We assume that → the contour γ(t) = u(t)+ iv(t), t [a, b], has continuously diﬀerentiable u and v on (a, b). ∈

We consider a subdivision

a = t0 <ξ0 < t1 <ξ1 < t2 < ... < ξn 1 < tn = b − of [a, b] and form the corresponding ”Riemann sum”

n 1 − f(γ(ξ ))[γ(t ) γ(t )] X. j j+1 − j ∈ j=0 X It can be proved that as the partition is reﬁned, this sum tends to a limit in X this limit is deﬁned to be γ f(z)dz. R Let X be a Banach space, let U be a starlike open set in C and let f : U X → be analytic. For any close contour γ in U and any point a U not on γ, ∈ f(z)dz =2πi n(γ; a)f(a) (4.2) z a Zγ − and f(z)dz =2πi n(γ; a)f ′(a). (4.3) (z a)2 Zγ − Lemma 4.4. Let X be a Banach space, let R > 0 and let f be an X-valued function analytic on the disc

(a, R)= z C : z a R . △ { ∈ | − | ≤ } Suppose f(z) M z (a, R). X ≤ ∀ ∈△ Then M f ′(a) . X ≤ R

19 Proof. In Cauchy’s Integral Formula (4.3), take γR to be the anti-clockwise oriented circle that bounds (a, R): △ γ (t)= a + Reit, 0 t 2π. R ≤ ≤ Then n(γR; a)=1, and so 1 f(z)dz f ′(a)= . 2πi (z a)2 ZγR − Hence 1 f(z)dz f ′(a) = . (4.4) X 2π (z a)2 ZγR X − The integral on the RHS is the limit of Riemann sums of the form

n 1 − f γ(ξ ) S = ◦ j (γ(t ) γ(t )) τ (γ(ξ ) a)2 j+1 − j j=0 j X − where τ : 0= t0 <ξ0 < t1 <ξ1 < t2 < ... < ξn 1 < tn =2π. − By the triangle inequality, n 1 − f γ(ξ ) S ◦ j X γ(t ) γ(t ) ≤ γ(ξ ) a 2 | j+1 − j | j=0 j X | − | M M 2πR =2π . ≤ R2 R On combining this inequality with (4.4) we have M f ′(a) . X ≤ R

Theorem 4.5. Liouville’s Theorem. A function that is analytic and bounded in the entire complex plane is constant. Proof. Let f : C X be analytic and bounded by M, so that → f(z) M X ≤ for all z C. Consider∈ any a C and R> 0. The function f is analytic on (a, R), and so, ∈ △ by Lemma 4.4, M f ′(a) . ≤ R Since this relation holds R > 0, f ′(a) X =0, and so f ′(a)=0. Thus f ′(z)=0 for all z C. It follows that∀ f is constant on C. ∈ 20 Theorem 4.6. For any element x in a Banach algebra A with identity e =0, the spectrum σ(x) of x is a non-empty closed bounded set in C.

Proof. By Theorem 3.8, σ(x) is closed and bounded in C. We must show σ(x) = . Suppose σ(x)= , so that ρ(x)= C. By Theorem 4.2, ∅ ∅ 1 x( ): C A : λ (λe x)− → → − is analytic on the entire complex plane. Note that x = 0 since 0 / σ(0). ∈ We claim that x( ) is a bounded function on C. If λ > 2 x , | | 1 1 1 x − (λe x)− ) = λ− e − − λ 2 1 x x = λ − e + + + ... (see Lemma 3.3) | | λ λ2 1 ∞ x n e + ≤ 2 x λ ! n=1 1 X ( e + 1). ≤ 2 x On the other hand x( ) is continuous on the closed bounded set z C : z 2 x , { ∈ | | ≤ } hence is bounded above by some M1 > 0. Let e +1 M = max M , . 1 2 x Then x(λ) M λ C. ≤ ∀ ∈ By Liouville’s Theorem 4.5, x( ) is constant 1 1 ∴ (λe x)− = x(0) = x− λ C − − ∀ ∈

∴ λe x = x λ C − − ∀ ∈

∴ λe =0 λ C, ∀ ∈ a contradiction to e = 0. Hence σ(x) = . ∅

21 5 The Gelfand-Mazur Theorem

Definition 5.1. A division algebra is a non-zero algebra with identity (over an arbitrary field) in which every non-zero element has a multiplicative inverse [is regular]. Example 5.2. Every field is a division algebra over itself. The quaternions H form a (non-commutative) division algebra over R. Definition 5.3. We say that two algebras A, B over C are isomorphic if there is a bijective mapping F : A B such that x, y A, λ C, → ∀ ∈ ∈ F (x + y)= F (x)+ F (y),

F (xy)= F (x)F (y),

F (λx)= λF (x). Deﬁnition 5.4. We say that two Banach algebras A, B over C are isomorphic if there is a bounded linear operator F : A B such that F is invertible and → x, y A, ∀ ∈ F (xy)= F (x)F (y). Theorem 5.5. Gelfand-Mazur Theorem. Every Banach division algebra over C is isomorphic the Banach algebra C. Proof. Let A be a Banach algebra with identity e =0, and suppose A is a division algebra over C.

Consider any x A. By Theorem 4.6, σ(x) = , so λ σ(x). Hence λe x is singular. Since A is∈ a division algebra, only 0 is singular.∅ ∃ ∈ −

∴ λe x =0, −

∴ x = λe. Deﬁne F : C A by F (λ)= λe. We have just shown that F is surjective. → Clearly F is injective, for if λ = and F (λ)= F () then F (λ) F ()=0. − ∴ (λ )e =0, −

1 ∴ (λ )− (λ )e =0, − − 22 ∴ e =0. This is contradiction. Hence F is bijective. For all λ, C, ∈ F (λ + )=(λ + )e = λe + e = F (λ)+ F (),

F (λ)=(λ)e =(λe)(e)= F (λ)F (),

F (λ)=(λ)e = (λe)= F (λ). Therefore F is an isomorphism of algebras over C. Note that F is bounded, since

F = sup λe = e . λ C, λ 1 ∈ | |≤ The inverse map is bounded too, since

1 F − : A C : x λ → → 1 where λ σ(x) and, by Theorem 3.8, λ x . Hence F − 1. Therefore F is an isomorphism∈ of Banach algebras A| |≤and C. ≤

23 6 Convolution algebras

The main examples of Banach algebras encountered so far are algebras of continuous functions like C[0, 1] and H∞(Ω), and algebras of operators like Mn(C) and B(X). Another important class comprises convolution algebras.

1 Example 6.1. The algebra ℓ (Z+). 1 Let Z+ be the additive semigroup of non-negative integers, and let ℓ (Z+) be the Banach space of sequences (x )∞ , with x C for all n 0, such that n n=0 n ∈ ≥

∞ x < , | n| ∞ n=0 X with co-ordinatewise addition and scalar multiplication and norm

∞ (x ) = x . n 1 | n| n=0 X 1 We could make ℓ (Z+) into a Banach algebra by introducing co-ordinatewise multiplication, but there is another natural multiplication which is more useful. n We can identify a sequence (xn)n 0 with the power series xnz , which we can ≥ think of as a formal power series or as the expansion of an analytic function on the unit disc (0, 1). Note that P △ 0 1 2 0 1 2 (x0z + x1z + x2z + ...)(y0z + y1z + y2z + ...)

0 1 2 = x0y0z +(x0y1 + x1y0)z +(x0y2 + x1y1 + x2y0)z n +... +(x0yn + x1yn 1 + ... + xny0)z + ... − This suggests the multiplication

(xn)n 0 (yn)n 0 =(x0yn + x1yn 1 + ... + xny0)n 0 ≥ ∗ ≥ − ≥ n

=( xjyn j)n 0 − ≥ j=0 X This is called the convolution multiplication.

24 1 If x =(xn)n 0, y =(yn)n 0 ℓ (Z+), then ≥ ≥ ∈ n

x y 1 = ( xjyn j) 1 ∗ − j=0 X n ∞ = xjyn j − n=0 j=0 X Xn ∞ xj yn j ≤ | || − | n=0 j=0 X X ∞ x y ≤ | j|| k| j,kX=0 = x y . 1 1 Thus . is submultiplicative on (ℓ1(Z ), +, ) and ℓ1(Z ) is a Banach algebra 1 + ∗ + with respect to . ∗

Example 6.2. The algebra ℓ1(Z). 1 Let ℓ (Z) be the Banach space of sequences (xn)n Z, xn C for all n Z, such ∈ that ∈ ∈ ∞ x < | n| ∞ n= X−∞ with co-ordinatewise addition and scalar multiplication and norm

∞ (xn)n Z 1 = xn . ∈ | | n= X−∞ n The sequence (xn)n Z can be identiﬁed with the Laurent series ∞ xnz . This ∈ n= power series does converge for z T, the unit circle, though we may−∞ think of it as a formal power series. Formally,∈ P

2 1 0 1 2 1 0 1 ... + x 2z− + x 1z− + x0z + x1z + ...... + y 2z− + y 1z− + y0z + y1z + ... − − − − ∞ ∞ 0 1 = + xky k z + xky1 k z + ... − − k= ! k= ! ! X−∞ X−∞ ∞ ∞ n = xkyn k z . − n= k= ! X−∞ X−∞

25 We deﬁne convolution multiplication on ℓ1(Z) by ∗ ∞ (xn)n Z (yn)n Z = xkyn k . ∈ ∗ ∈ − k= !n Z X−∞ ∈ As in the previous example, we ﬁnd, for x, y ℓ1(Z), ∈ x y x y . ∗ 1 ≤ 1 1 Hence ℓ1(Z) is a commutative Banach algebra with respect to . Since ℓ1(Z) can be n ∗ thought of as an algebra of power series ∞ xnz with pointwise multiplication, it is clear that it has an identity −∞ P e =(...0, 0, 1, 0, 0, ...), where 1 is in the ”zeroth co-ordinate”, corresponds to the constant function 1.

Example 6.3. Group algebras CG. Let G be a ﬁnite group. The group algebra CG is the algebra of formal sums

g G agg where ag C, with the natural addition and scalar multiplication, and ∈ ∈ multiplication given by P ∗

a g b g = a b g . g ∗ g k h g G ! g G ! g G kh=g ! ! X∈ X∈ X∈ X

The formal sum g G agg is really ”the same thing” as the function a : G C ∈ → given by a(g)= ag. With this notation, a b is given by P ∗ a b : G C : g a(k)b(h) ∗ → → k,h G, kh=g ∈X 1 = a(k)b(k− g) k G X∈ 1 = a(gh− )b(h). h G X∈ Hence CG is a ﬁnite dimensional algebra with identity 1eG, where eG is the identity of G. Note that C is commutative G is abelian. The algebra C is a Banach G ⇐⇒ G algebra with respect to the norm

a g = a . g 1 | g| g G g G X∈ X∈

26 Example 6.4. The algebra L1(R). Let L1(R) be the Banach space of Lebesgue measurable functions f : R C such that → ∞ f(t) dt < | | ∞ Z−∞ with pointwise addition and scalar multiplication and norm

∞ f = f(t) dt. 1 | | Z−∞ In this space we regard two functions f and g as equal if they agree ”almost everywhere”, that is, everywhere except a countable sequence of points. One of the most signiﬁcant properties of the Lebesgue integral is that L1(R) is complete.

For f,g L1(R) we deﬁne f g : R C by ∈ ∗ → ∞ f g(t)= f(s)g(t s)ds. ∗ − Z−∞ We shall show that f g L1(R) using two properties of the Lebesgue integral. ∗ ∈ 1. Translation invariance. For h L1(R), s R, ∈ ∈ ∞ ∞ h(t s)dt = h(t)dt. − Z−∞ Z−∞

2. The Fubini-Tonnelli theorem. For an integrable Lebesgue measurable function F : R R R, if F (s, t) 0 for all s, t R, × → ≥ ∈ ∞ ∞ ∞ ∞ F (s, t)dtds = F (s, t)dsdt Z−∞ Z−∞ Z−∞ Z−∞ (”reversing the order of integration”).

27 We have

∞ ∞ ∞ (f g)(t) dt = f(s)g(t s)ds dt | ∗ | − Z−∞ Z−∞ Z−∞ ∞ ∞ f(s) g(t s) dsdt ≤ | || − | Z−∞ Z−∞ ∞ ∞ = f(s) g(t s) dtds (the Fubini-Tonnelli theorem) | || − | Z−∞ Z−∞ ∞ ∞ = f(s) g(t s) dt ds | | | − | Z−∞ Z−∞ ∞ ∞ = f(s) g(t) dt ds (translation invariance) | | | | Z−∞ Z−∞ ∞ = f(s) g ds | | 1 Z−∞ = f g 1 1 Hence f g L1(R), and, moreover, . is submultiplicative. ∗ ∈ 1 Let us show that is associative. For f,g,h L1(R) and t R, ∗ ∈ ∈ ∞ ((f g) h)(t)= (f g)(s) h(t s) ds ∗ ∗ ∗ − Z−∞ ∞ ∞ = f(u)g(s u)du h(t s)ds − − Z−∞ Z−∞ ∞ ∞ = f(u)g(s u)h(t s) ds du (the Fubini-Tonnelli theorem) − − Z−∞ Z−∞ ∞ ∞ = f(u) g(s u)h(t s) ds du − − Z−∞ Z−∞ ∞ ∞ = f(u) g(s)h(t u s)ds du (translation invariance) − − Z−∞ Z−∞ ∞ = f(u)(g h)(t u) du ∗ − Z−∞ =(f (g h))(t). ∗ ∗ Hence is associative. Thus L1(R) is a Banach algebra. It is commutative, but ∗ has no identity.

Note the probabilistic interpretation of : if f,g are probability density functions of random variables X and Y then f ∗ g is the probability density function of X + Y. ∗

28 7 Homomorphisms and ideals

Deﬁnition 7.1. Let A, B be Banach algebras. A homomorphism from A to B is a bounded linear operator φ : A B → such that, x, y A, ∀ ∈ φ(xy)= φ(x)φ(y). If A, B have identities e , e respectively and a homomorphism φ : A B satisﬁes A B →

φ(eA)= eB then φ is said to be a unital homomorphism. Examples 7.2. 1. For any Banach algebras A and B, the map φ : A B : x 0 is a homomorphism, called the zero homomorphism. → → 2. For α [0, 1], let ∈ v : C[0, 1] C : x x(α). α → → The map v is ”evaluation at α”. For x, y C[0, 1], and λ C, α ∈ ∈ vα(x + y)=(x + y)(α)= x(α)+ y(α)= vα(x)+ vα(y), vα(λx)=(λx)(α)= λx(α)= λvα(x), vα(xy)=(xy)(α)= x(α)y(α)= vα(x)vα(y). The operator norm

vα = sup vα(x) = sup x(α) =1. x ∞ 1 | | x ∞ 1 | | ≤ ≤ Recall that C[0, 1] has an identity e given by e(t)=1 t [0, 1]. Thus ∀ ∈ vα(e)= e(α) = 1 the identity of C. Hence vα is a unital homomorphism of Banach algebras.

3. Notation. If X,Y are sets, M X and f : X Y is a map then f M is the restriction of f to M, that is,⊂ the map → |

f : M Y : x f(x). |M → → Deﬁne 1 φ : C[0, 1] C 0, : f f [0, 1 ]. → 2 → | 2 It is easy to check that φ 1 ≤ and φ is a homomorphism.

29 A 0 4. φ : M (C) M (C): A is a homomorphism. 2 → 4 → 0 A 5. For α (0, 1) deﬁne ∈△ ∞ 1 n vα : ℓ (Z+) C :(xn)n Z+ xnα . → ∈ → n=0 X Then vα is a homomorphism. Linearity is easy.

∞ v (x y)= (x y) αn α ∗ ∗ n n=0 X n ∞ n = xkyn kα − n=0 X Xk=0 ∞ n+m = xnymα n,m=0 X ∞ ∞ n m = xnα ymα n=0 m=0 X X = vα(x)vα(y). See Example 6.1. 6. Let C( (0, 1)) be the Banach algebra of continuous C-valued functions on △ (0, 1) with sup norm. Deﬁne △ φ : ℓ1(Z ) C( (0, 1)) + → △ 1 by if x =(xn)n 0 ℓ (Z+), φ(x) is the function ≥ ∈ ∞ z x zn in C( (0, 1)). → n △ 0 X The proof that φ(x y)= φ(x)φ(y) is the same calculation as in (5). All except (1) are unital homomorphisms.∗ None of them are isomorphisms, i.e. none is bijective. All of them are continuous operators. Deﬁnition 7.3. A character of a commutative Banach algebra A is a non-zero homomorphism from A to C. Examples 7.2 (2) and (5) above are examples of characters.

30 Remark 7.4. φ(A) is a subspace of C and is not 0 , hence is C. Thus φ is surjective. { } Theorem 7.5. Let φ be a character of a commutative Banach algebra A with identity e =0. Then φ(e)=1 and, for any x A, ∈ φ(x) σ(x), ∈ and so φ(x) x . | |≤ Proof. Since φ(A)= C there exists a A such that φ(a)=1. Then ∈ 1= φ(a)= φ(ae)= φ(a)φ(e)= φ(e). Hence φ is unital homomorphism.

Suppose φ(x) / σ(x), that is, φ(x)e x is a regular element in A. Hence it has ∈ − an inverse u A: ∈ (φ(x)e x)u = e. − Since φ is multiplicative, φ((φ(x)e x)u)= φ(e)=1, − ∴ (φ(x)φ(e) φ(x))φ(u)=1, − ∴ 0 = 1 - contradiction. Hence φ(x) σ(x). By Theorem 3.8, ∈ φ(x) x . | |≤

Corollary 7.6. Let A be a commutative Banach algebra with identity e = 0. Characters of A are bounded linear functionals on A, and so belong to the dual space A∗ of A and have norm 1. ≤ Characters are also called multiplicative linear functionals.

Deﬁnition 7.7. The set of characters of a commutative Banach algebra A is called the character space of A and is denoted by A.ˆ Examples 7.8. (1) C[0, 1] v :0 α 1 . ⊃{ α ≤ ≤ } (2) ℓ1(Z ) φ : α (0, 1) . + ⊃{ α ∈△ } Does equality hold? To be answered later.

31 Remark 7.9. Suppose φ : A B is a homomorphism of algebras over C. Recall that → ker φ = x A : φ(x)=0 . { ∈ } Kerφ is also called the null space of φ.

Since φ is a linear mapping, Ker φ is a linear subspace of A.

If x ker φ and a A then ∈ ∈ φ(ax)= φ(a)φ(x)= φ(a)0=0, and similarly φ(xa)=0.

Hence if x ker φ, then ax ker φ and xa ker φ for all a A. ∈ ∈ ∈ ∈ Deﬁnition 7.10. An ideal in an algebra A is a linear subspace I of A such that, for any x I and a A, ax I and xa I. An closed∈ ideal in a Banach∈ algebra∈ A is∈ a closed linear subspace I of A such that, for any x I and a A, ax I and xa I. ∈ ∈ ∈ ∈ An ideal I is proper if I = A. Examples 7.11. 1. In any algebra A, 0 and A are ideals. { } 2. In C[0, 1], for any set E [0, 1], ⊂ x C[0, 1] : x E =0 { ∈ | } is an ideal.

In particular, for α [0, 1], ∈ ker v = x C[0, 1] : x(α)=0 α { ∈ } is an ideal.

3. In ℓ1(Z ), for α (0, 1), + ∈△

∞ ker φ = (x ) ℓ1(Z ): x αn =0 α { n ∈ + n } n=0 X is an ideal.

32 4. In C[z], f : f(3 i)= f(4+2i)=0 { − } is an ideal.

5. c0 is an ideal in ℓ∞. Theorem 7.12. In an algebra A with identity e = 0, let I be an ideal. The following are equivalent:

1. I is proper;

2. e / I; ∈ 3. I does not contain any regular element.

Proof. (3) (2) (1) is obvious. ⇒ ⇒ To prove (1) (3), suppose I is proper and I contains a regular element x. ⇒ 1 Then x has an inverse x− A, and so for any a A we have ∈ ∈ 1 (ax− )x I. ∈ Therefore, a I. Hence A = I, this is a contradiction. Thus (1) (3). ∈ ⇒ Consequently (1)-(3) are equivalent.

Deﬁnition 7.13. An ideal I in an algebra A is maximal if it is proper and the only ideals of A which contain I are I and A.

Examples 7.14. 1. 0 is a maximal ideal in C. { } 2. In C[0, 1], I = f : f(0) = 0 = f(1) is an ideal, but is not maximal since { } the ideal J = f : f(0) = 0 { } is an ideal, J I and J is not equal to I or C[0, 1]. ⊃ Theorem 7.15. If φ is a character of a commutative Banach algebra A then ker φ is a maximal ideal of A.

Proof. Let M = ker φ. Clearly M is an ideal of A, and M is proper, else φ is a zero homomorphism contrary to the deﬁnition of character.

Let us show that M is maximal. Suppose M I A, where I is an ideal of A and M = I. We wish to show I = A. Observe that⊂ ⊂φ(I) is a linear subspace of 33 C, and hence either φ(I)= 0 or φ(I)= C. { } If φ(I) = 0 then I ker φ = M, hence I = M, contrary to hypothesis. { } ⊂ Hence φ(I)= C.

Pick x I such that φ(x)=1. For any a A we have ∈ ∈ φ(a φ(a)x)= φ(a) φ(a)=0, − − and so a φ(a)x M. − ∈ Hence a φ(a)x + M I + M = I. ∈ ∈ That is, a I. Hence I = A. Thus M is a maximal ideal. ∈ 1 Example 7.16. ker v 1 i = f H∞(D): f( i)=0 is a maximum ideal in 2 { ∈ 2 } H∞(D).

Theorem 7.17. In an algebra A with identity every proper ideal is contained in a maximal ideal.

Proof. Let I be an ideal of A and let e be the identity element of A.

Let be the set of proper ideals of A that contain I. is a partially ordered J J set under inclusion: means . J1 ≤J2 J1 ⊂J2 If is a chain in then J J is an ideal of A containing I, and J J is proper C J ∈C ∈C since it does not contain e. That is, J J , and so every chain in has an upper bound in . By Zorn’sS Lemma, ∈C contains∈J a maximal elementS JM. Thus J J M is a proper ideal of A containingSI. If J is an ideal of A containing M then either J = A or J , and hence, since M is a maximal element of , J = M. Therefore M is a maximal∈ J ideal of A, and I M. J ⊂

Corollary 7.18. Let A be a commutative algebra with identity. For an element x A, x is singular if and only if x belongs to some maximal ideal of A. ∈ Proof. If x is not singular then, by Theorem 7.12, x does not belong to any proper⇐ ideal of A, and hence x does not belong to any maximal ideal. Therefore, x is in some maximal ideal that x is singular. ⇒

34 Suppose x is singular. Let ⇒ xA = xa : a A . { ∈ } The linear subspace xA is an ideal of A and, since x is singular, e / xA. Hence, by Theorem 7.12, xA is a proper ideal, and x = xe xA. By Theorem∈ 7.17, there is a maximal ideal M of A such that xA M. Hence∈ x belongs to some maximal ⊂ ideal of A.

35 8 The algebra C(K).

Definition 8.1. Let K be a compact Hausdorff space. C(K) denotes the Ba- nach algebra of continuous C-valued functions on K with pointwise operations and supremum norm f = supt K f(t) . Thus, for f,g C(K) and λ C, we ∞ ∈ | | ∈ ∈ define (f + g)(t)= f(t)+ g(t), t K, ∈ (λf)(t)= λf(t), t K, ∈ (fg)(t)= f(t)g(t), t K. ∈

Theorem 8.2. Let K be a compact Hausdorﬀ space. Then (C(K), ) is a ∞ Banach algebra. Proof. For f C(K), ∈ sup f(t) < t K | | ∞ ∈ since continuous R-valued functions on compact sets are bounded. Hence is ∞ well deﬁned on C(K).

The linear space C(K) is closed under the stated algebraic operations. For example, if f,g C(K) then f + g is continuous, being the composition of two continuous maps:∈ K C C C : → × → t (f(t),g(t)) f(t)+ g(t). → → It is straightforward to check that C(K) is a normed algebra over C. For all f,g C(K), ∈ fg = sup f(t)g(t) = sup f(t) g(t) sup f(t) sup g(t) = f g . ∞ t K | | t K | || | ≤ t K | | t K | | ∞ ∞ ∈ ∈ ∈ ∈ Completeness of (C(K), ). Let (fn)n∞=1 be a Cauchy sequence in C(K). ∞ We must show (fn)n∞=1 tends to a limit f in (C(K), ). ∞

For any t K, (f (t))∞ is a Cauchy sequence in C since ∈ n n=1 fn(t) fm(t) fn fm . | − |≤ − ∞ The normed space (C, ) is complete, and so there exists f(t) C such that || ∈ lim fn(t)= f(t). n →∞ We will show below that f is continuous on K and that fn f 0 as ∞ n . − → →∞ 36 Digression on limits of sequences of continuous functions.

It is not true in general that if fn is continuous on K for every n and fn(t) f(t) as n for all t K then f is continuous on K. → →∞ ∈

Example 8.3. Let

1 nt if 0 t 1/n, fn(t)= − ≤ ≤ (0 if t> 1/n.

1 if t =0, Then fn(t) f(t) as n , where f(t) = is not continuous → → ∞ (0 if t> 0 on [0, 1].

Deﬁnition 8.4. Let fn, n N, and g be C-valued functions on a set E. We say that f g ∈ n →

(i) pointwise on E as n if limn fn(t)= g(t) for every t E; →∞ →∞ ∈ (ii) uniformly on E as n if, for every ǫ> 0, there exists N N such that n N implies →∞ ∈ ≥ f (t) g(t) <ǫ for all t E. | n − | ∈

In the above example fn f as n pointwise but not uniformly. The example shows that a pointwise→ limit of continuous→ ∞ functions on [0, 1] need not be continuous.

Theorem 8.5. If (fn)n∞=1 is a sequence of continuous C-valued maps on a topological space E and fn g uniformly on E as n then g is continuous on E. → → ∞

ǫ Proof. This is an ” 3 argument”. Consider t E. We want to show that g is continuous at t. ∈ Let ǫ> 0. Since f g uniformly there exists N N such that n N implies n → ∈ ≥ f (s) g(s) < ǫ for all s E. | n − | 3 ∈ Since fN is continuous at t there is a neighbourhood U of t in E such that ǫ f (t) f (s) < for all s U. | N − N | 3 ∈

37 Then, for any s U, ∈ g(t) g(s) = g(t) f (t)+ f (t) f (s)+ f (s) g(s) | − | | − N N − N N − | g(t) fN (t) + fN (t) fN (s) + fN (s) g(s) ≤|ǫ ǫ− ǫ | | − | | − | < + + 3 3 3 = ǫ.

Hence g is continuous at t E. ∈ Return to the proof that C(K) is complete (Theorem 8.2).

Recall that (f )∞ is a Cauchy sequence and f (t) f(t) as n for every n n=1 n → →∞ t K. ∈ Let ǫ> 0. There exists N N such that m, n N implies ∈ ≥

fm fn < ǫ, − ∞ and hence, for any t K, ∈ f (t) f (t) < ǫ. | m − n | Therefore lim fm(t) fn(t) ǫ m →∞ | − | ≤ and f (t) f(t) ǫ t K and n N. | n − | ≤ ∀ ∈ ≥ Thus fn f uniformly on K as n . By Theorem 8.5, f is continuous on K. Moreover,→ for any ǫ> 0, N N such→∞ that ∃ ∈

fn f ǫ whenever n N, − ∞ ≤ ≥ by above. Thus fn f as n in (C(K), ). → →∞ ∞ Hence C(K) is complete with respect to . Thus C(K) is a Banach algebra. ∞ Note that C(K) is a commutative Banach algebra with identity e where e(t) = 1 t K. ∀ ∈

38 9 Characters of C(K)

Deﬁnition 9.1. For t K, we call the functional ∈ v : C(K) C : f f(t) t → → an evaluation functional on C(K).

Evaluation functionals are characters of C(K).

Theorem 9.2. Let K be a compact Hausdorﬀ space. Let φ be a character of C(K). There exists t K such that 0 ∈ φ(f)= f(t ) f C(K) 0 ∀ ∈ Thus the characters of C(K) are precisely the evaluation functionals.

Proof. Let M = ker φ. Since φ is non-zero, M is a proper ideal in C(K). Suppose there is no t K such that every element of M vanishes at t . Then, 0 ∈ 0 for each t K, there exists ft M such that ft(t) =0. Let g =∈ f¯ f : then g M ∈and g 0 on K and g (t) > 0. Let t t t t ∈ t ≥ t 1 U = g− (0, )= τ K : g (τ) > 0 . t t ∞ { ∈ t }

Then Ut is an open neighbourhood of t, and so Ut : t K is an open cover of K. Since K is compact, there exists a ﬁnite subcover{ ∈ U }, ..., U for some { t1 tn } t , ..., t K. Let 1 n ∈ g = gt1 + gt2 + ... + gtn . Then g M and g > 0 on K. Hence 1 C(K), that is, g is a regular element of ∈ g ∈ C(K). This contradicts the fact that M is a proper ideal of C(K) (see Theorem 7.12). Hence t K such that ∃ 0 ∈ f(t )=0 f M. 0 ∀ ∈ In other words, M f C(K): f(t )=0 . ⊂{ ∈ 0 } By Theorem 7.15, M is a maximal ideal of C(K). Thus, by Theorem 7.17, we must have M = f C(K): f(t )=0 . { ∈ 0 }

39 Since φ is non-zero, we have φ(e) =0. However, since e2 = e, we have φ(e)2 = φ(e), and so φ(e)=1. For any h C(K ), we can write ∈ h =(h h(t )e)+ h(t )e − 0 0 where h h(t )e vanishes at t hence belongs to M. Thus − 0 0

φ(h)=0+ h(t0)φ(e)= h(t0).

Consequently φ is evaluation at t0. We can state this result:

C(K)Λ = v : t K { t ∈ } where v is evaluation at t, that is, v (f)= f(t) for all f C(K). t t ∈

40 Part II. Topics in Topology

This part concerns the theory of pseudometrics on sets, weak∗-topologies on dual Banach spaces and product topologies. We shall deﬁne these objects and study their properties.

Contents of Part II

Pages 10. Metric spaces 42

11. Topologies of metric spaces 44

12. Topological spaces 45

13. Pseudometrics 47

14. Hausdorﬀ spaces 49

15. Compactness 50

12. Topological subspaces 53

17. Product topologies 55

18. Inﬁnite products 58

41 10 Metric spaces

Deﬁnition 10.1. A metric space is a pair (M, d) where M is a set and a metric

d : M M R+ × → satisﬁes

1. d(x, y)=0 x = y ⇐⇒ 2. d(x, y)= d(y, x) (symmetry)

3. d(x, z) d(x, y)+ d(y, z) (triangle inequality) ≤ for all x, y, z M. ∈ Examples 10.2. 1. Any normed linear space X is a metric space with respect to the metric d(x, y)= x y for all x, y X. − ∈ 2. Any subset of (X, d) of a metric space (M, d) is a metric space with respect to the restriction of d, since the restriction of a metric is a metric.

3. For any set M, deﬁne d : M M R+ × → by 0 if x = y d(x, y)= 1 if x = y. ( This is the “small world” metric.

Deﬁnition 10.3. In a metric space (M,d) we deﬁne, for ǫ > 0, the open ball of centre x, radius ǫ, to be

B (x)= y M : d(x, y) < ǫ M. ǫ { ∈ } ⊂ For any set U M we say that x U is an interior point of U if there exists ǫ> 0 such that B ⊂(x) U. ∈ ǫ ⊂ We say that U is an d-open set in M if every point of U is an interior point of U.

Thus U is an open set in (M,d) if for all x U there exists ǫ > 0 such that ∈ B (x) U. If we need to emphasize the metric we write B (x, d). ǫ ⊂ ǫ

42 Examples 10.4. 1. For (M,d) as in the discrete metric, B 1 (x) = x for any 2 x M. Thus every subset of M is open. ∈ 2 2. Let M = R and let d1,d2,d be the metrics induced by the norms . 1, . 2 ∞ and . respectively. Thus ∞ (x , x ) = x + x 1 2 1 | 1| | 2| 2 2 1 (x , x ) = x + x 2 1 2 2 {| 1| | 2| } (x1, x2) = max x1 , x2 , ∞ {| | | |} and for x, y C2, ∈ d (x, y)= x y 1 − 1 d (x, y)= x y 2 − 2 d (x, y)= x y . ∞ − ∞ The set U = (x , x ): x > 0, x > 0 is open with respect to all 3 metrics. { 1 2 1 2 } Choose ǫ = min x , x > 0. Then { 1 2}

Bǫ(x, d1) U, Bǫ(x, d2) U, Bǫ(x, d ) U. ⊂ ⊂ ∞ ⊂

43 11 Topologies of metric spaces

Definition 11.1. Two metrics on a set M are said to be topologically equivalent if they give rise to the same open sets. 2 Note that d2 and d are topologically equivalent metrics on R , while d2 and ∞ d defined by 0 if x = y d(x, y)= 1 if x = y, ( are topologically inequivalent, since the singleton set 0 is d-open, but not d2- open. { } Definition 11.2. The topology of a metric space (M,d) is the collection of d-open sets in M. J Theorem 11.3. The topology of a metric space (M,d) satisfies the following conditions: J T1. and M ; ∅∈J ∈J T2. if U, V then U V ; ∈J ∩ ∈J T3. if U : α A is a collection of members of , for any index set A, then { α ∈ } J U . α ∈J α A [∈ Proof. T1. It is immediate. T2. Let U, V and let x U V. Choose ǫ , ǫ such that B (x) U, ∈ J ∈ ∩ 1 2 ǫ1 ⊂ Bǫ2 (x) V. Let ǫ = min ǫ1, ǫ2 . Then Bǫ(x) U V. Hence x is interior to U V. ⊂Thus U V .{ } ⊂ ∩ ∩ ∩ ∈J

T3. Let Uα for α A and let U = α Uα. Consider x U. Then x Uβ for some β ∈JA, and hence∈ there exists ǫ> 0 such that B ∈(x) U U.∈ Hence ∈ ǫ ⊂ β ⊂ x is interior to U. Thus U . S ∈J

Remark 11.4. It follows from (T2) that the topology of a metric space is closed under ﬁnite intersections: if U , U , ..., U are open setsJ then U ... U is 1 2 n 1 ∩ ∩ n open. However is not in general closed under inﬁnite intersections. In R, with its standard metricJ (d(x, y)= x y ), take | − | 1 1 Un =( , )= B 1 (0). −n n n U U ... = 0 which is not open. 1 ∩ 2 ∩ { }

44 12 Topological spaces

In analysis we often need a more subtle notion of “nearness” or continuity than that provided by a metric. The idea of a topology provides such a notion.

Deﬁnition 12.1. A topology on a set X is a family of subsets of X with the J properties:

T1. , X ; ∅∈J ∈J T2. is closed under ﬁnite intersections; J T3. is closed under arbitrary unions. J

A topological space is a pair (X, ) where X is a set and is a topology on X. Elements of are called open setsJ or -open sets. J J J By Theorem 11.3, every metric space is naturally a topological space.

Not all topological spaces are obtainable from metrics.

Examples 12.2. 1. Let (M,d) is a metric space and let be the collection of d-open sets. Then (M, ) is a topological space. J J 2. Let X be a set. The indiscrete or trivial topology on X is = ,X . Thus an indiscrete topological space is J {∅ }

(X, ) where = , X . J J {∅ } 3. The discrete topology: (X, ) where is the collection of all subsets of X. J J

0 if x = y The topology is induced by the ”small world metric”: d(x, y)= 1 if x = y. ( 4. The coﬁnite topology: (N, ) where consists of the empty set and all sets with ﬁnite complement inJN. J

5. Rn, Cn have a natural or standard topology, induced by the Euclidean norm.

45 Remark 12.3. Note that in a metric space (M,d), for any x M the set ∈ U = M x is an open set. For y U such that d(x, y) > 0, we have Bǫ(y) U \{1 } ∈ ⊂ where ǫ = 2 d(x, y).

In Example 2 with X having more than one point, the complements of single- tons do not belong to . Hence there is no metric d on X such that the topology J corresponding to d is . J Thus the notion of a topological space is more general than the notion of a metric space.

46 13 Pseudometrics

Deﬁnition 13.1. A pseudometric on a set X is a function

d : X X R+ × → which satisﬁes, for all x, y, z X, ∈ 1. d(x, y)= d(y, x);

2. d(x, z) d(x, y)+ d(y, z). ≤ Note: it can happen that d(x, y) = 0 even though x = y. Example 13.2. On C[0, 1], define d , for 0 t 1, by t ≤ ≤ d (f,g)= f(t) g(t) . t | − | Note that d is a pseudometric for t [0, 1]. t ∈ Definition 13.3. Let be a non-empty set of pseudometrics on a set X. For any x X,ǫ > 0 and any finiteD set d , ..., d , we define ∈ 1 n ∈D B (x; d , ..., d )= y X : d (x, y) < ǫ for j =1, 2, ..., n . ǫ 1 n { ∈ j } For any subset U of X, we say that x is an interior point of U with respect to if there exist ǫ> 0 and finitely many d , ..., d such that D 1 n ∈D B (x; d , ..., d ) U. ǫ 1 n ⊂ A set U X is -open if every point of U is interior with respect to . ⊂ D D Lemma 13.4. B (x, d , ..., d ) is -open. ǫ 1 n D Proof. Consider y B (x, d , ..., d ), so that ∈ ǫ 1 n

dj(x, y) < ǫ, for all j =1, 2, . . . , n.

Let η = min (ǫ dj(x, y)). 1 j n ≤ ≤ − Then η > 0, and B (y,d , ..., d ) B (x, d , ..., d ). η 1 n ⊂ ǫ 1 n Thus y is an interior point of B (x, d , ..., d ), and B (x, d , ..., d ) is -open. ǫ 1 n ǫ 1 n D

47 Theorem 13.5. Let be a set of pseudometrics on a set X. The collection of -open sets is a topologyD on X. D Proof. Let be the collection of -open subsets of X. J D T1. ,X trivially. ∅ ∈J T2. Suppose U, V are -open and consider x U V. There exist ǫ1, ǫ2 > 0 and d , ..., d ,d , ..., dD such that ∈ ∩ 1 n n+1 m ∈D B (x; d , ..., d ) U, B (x; d , ..., d ) V. ǫ1 1 n ⊂ ǫ2 n+1 m ⊂ Then, if ǫ = min ǫ , ǫ , { 1 2} B (x; d , ..., d , ..., d ) U V. ǫ 1 n m ⊂ ∩ Thus every point of U V is -interior to U V , that is, U V . ∩ D ∩ ∩ ∈J T3. Suppose U α Λ, and consider α ∈ J ∀ ∈ x U = U . ∈ α α Λ [∈ We have x U for some β Λ, and there exist ǫ > 0, d , ..., d such ∈ β ∈ 1 n ∈ D that B = Bǫ(x; d1, ..., dn) Uβ. Clearly B U, and so x is -interior to U. Thus U is -open. ⊂ ⊂ D D

Deﬁnition 13.6. The collection of -open sets is called the topology induced by the set of the set of pseudometrics.J D D Examples 13.7. 1. On C[0, 1], let = d :0 t 1 D { t ≤ ≤ } where d (f,g)= f(t) g(t) . t | − | For t , ..., t [0, 1], 1 n ∈ B (f,d , ..., d )= g C[0, 1] : f(t ) g(t ) < ǫ, 1 j n . ǫ t1 tn { ∈ | j − j | ≤ ≤ } The topology induced by is called the topology of pointwise convergence D on C[0, 1].

2. Let E be a Banach space and E∗ its dual. For any x E, deﬁne a pseudo- ∈ metric dx on E∗ by

d (F,G)= F (x) G(x) , F,G E∗. x | − | ∈ The set d : x E of pseudometrics induces a topology on E∗ called the { x ∈ } w∗-topology.

48 14 Hausdorﬀ spaces

Deﬁnition 14.1. A topological space X is Hausdorﬀ if, for any pair x, y of distinct points in X, there exists disjoint open sets U, V in X containing x, y respectively.

Examples 14.2. 1. Metric space are Hausdorﬀ.

2. In N with the cofinite topology, any pair of non-empty open sets have (in- finitely many) points in common and so the space is not Hausdorff.

3. If E is a Banach space and E∗ is its dual space, then E∗ is a Hausdorﬀ space in the w∗-topology.

Consider distinct elements F,G E∗. Since F = G, there exists x E such ∈ ∈ that F (x) = G(x). Let 1 1 ǫ = F (x) G(x) = d (F,G). 2| − | 2 x Then ǫ> 0, and Bǫ(F ; dx), Bǫ(G; dx)

are disjoint open sets containing F,G respectively. Therefore E∗ is Hausdorﬀ.

Note: Since N with the coﬁnite topology is not Hausdorﬀ, it’s topology does not arise from a metric.

Note: Not all Hausdorﬀ topologies arise from metrics. In fact E∗ with the w∗-topology (with dim E = ) does not. ∞

Deﬁnition 14.3. For any (x )∞ in a topological space (X, ) we say x a as n n=1 J n → n , or →∞ lim xn = a, n →∞ if, for every neighbourhood U of a, there exists N N such that x U for all ∈ n ∈ n N. ≥

In C[0, 1], with the topology of pointwise convergence,

f g f (t) g(t) t [0, 1]. n → ⇐⇒ n → ∀ ∈

49 15 Compactness

Deﬁnition 15.1. Let (X, ) be a topological space. A cover of a subset Y of X is a collection of subsets ofJX whose union contains Y.

Examples 15.2. 1. In R, a cover of [0, 1] is ( 1, 3 ), ( 1 , 2) . This is a ﬁnite { − 4 4 } cover.

2. In R, a cover of (0, 1) is the collection of sets U : x (0, 1) where { x ∈ } x 1+ x U =( , ). x 2 2

Note that no finite subcollection of the Ux covers (0, 1) : if x1 < x2 < ... < xn, then x 1+ x U ... U ( 1 , n ) x1 ∪ ∪ xn ⊂ 2 2 and so U ... U is a proper subset of (0, 1). x1 ∪ ∪ xn Definition 15.3. Let (X, ) be a topological space. An open cover of a subset Y of X is a cover of Y consistingJ of open sets. A subcover of a cover of Y is a U subcollection of that covers Y . U Definition 15.4. Let (X, ) be a topological space. A subset Y of X is compact J if every open cover of Y has a finite subcover.

Theorem 15.5. Any bounded closed interval in R is compact.

Theorem 15.6. Let (X, ) be a compact topological space. Any continuous func- J tion f : X R is bounded and attains its supremum and inﬁmum on X. → 1 Proof. Let U = f − ( n, n), since f is continuous, U is open, and U : n N n − n { n ∈ } is an open cover of X. It therefore has a ﬁnite subcover U , ..., U , where { n1 nk } n1 < ... < nk. Thus f(x) < nk x X, and so f is bounded. Let | | ∀ ∈ M = sup f(x) < . x X ∞ ∈ Suppose f does not attain its supremum M on X: this means that f(x) < M x X. For each x X pick y such that ∀ ∈ ∈ x

f(x)

50 Ux is an open neighbourhood of x, since f is continuous. The family Ux : x X is an open cover of X, and so by compactness it has an open subcover:{ ∈ } X U ... U ⊂ x1 ∪ ∪ xn for some x , ..., x X. 1 n ∈ We have f(x) y on U , and so ≤ xj xj

sup f(x) max yxj < M, x X ≤ 1 j n ∈ ≤ ≤ contrary to choice of M as the least upper bound of f. Hence f attains its supremum on X.

Similarly f attains its infimum on X. Theorem 15.7. A closed subset of a compact topological space is compact. Proof. Let (X, ) be compact and let Y be closed in X. Then X Y is open. J \ Consider any open cover of Y . Then ′ = U : U and X Y is an open U U { ∈ U \ } cover of X. Since X is compact, ′ has a finite subcollection that covers X: U X =(X Y ) U ... U , \ ∪ 1 ∪ ∪ n some U1, ..., Un . Thus ∈ U Y U ... U . ⊂ 1 ∪ ∪ n Hence has a finite subcover. Therefore Y is compact. U Theorem 15.8. A compact subset of a Hausdorff space is closed. Proof. Let (X, ) be a Hausdorff space and let Y be a compact subset of X. Consider any xJ X Y. For each y Y, by the Hausdorff condition, there exist ∈ \ ∈ disjoint open neighbourhoods Uy of y, Vy of x.

The family U : y Y is an open cover of the compact Y , hence has a ﬁnite { y ∈ } subcover U1, ..., Uyn with yj Y. Let V (x)= Vy1 ... Vyn , then V (x) is an open neighbourhood{ of x.}V (x) is disjoint∈ from U ...∩ U∩ , hence from Y , so that y1 ∪ ∪ yn V (x) X Y. ⊂ \ We have X Y = V (x), \ x Y X ∈[\ and since each V (x) is open in X, it follows that X Y is open in X. Hence Y is closed. \

51 Examples 15.9. Non-compact spaces.

x 1+x 1. (0, 1) is non-compact. ( 2 , 2 ): x (0, 1) is an open cover having no ﬁnite subcover. { ∈ }

2. R is non-compact. ( n, n): n N is an open cover having no ﬁnite subcover. { − ∈ }

3. The closed unit ball in ℓ2 is non-compact in the norm topology.

Theorem 15.10 (Heine-Borel). Any closed bounded set in Rn is compact.

Theorem 15.11. For any normed linear space E, the closed unit ball of E∗ is compact in the w -topology. That is, F E∗ : F 1 is compact in the topology deﬁned by∗ the pseudometrics d{ : x∈ E , where ≤ } { x ∈ } d (F,G)= F (x) G(x) . x | − | One can ﬁnd a proof of this theorem in Walter Rudin’s book “Functional Analysis”, 1973, Theorem 3.15.

52 16 Topological subspaces

Deﬁnition 16.1. Let (X, ) be a topological space and let Y be a subset of X. The relative topology of Y inJ X, or the induced topology of Y , is the topology

= U Y : U . JY { ∩ ∈J} Check that is a topology on Y . JY There are two natural ways to speciﬁy a topology on Q : by the natural metric d(x, y) = x y , x,y Q, or by giving Q the relative topology as a subset of R (with its natural| − | metric).∈ In fact they coincide. Theorem 16.2. Let be a set of pseudometrics which deﬁnes a topology on D J a set X, and let Y be a subset of X. The induced topology Y coincides with the topology determined by the pseudometrics J

Y = d Y Y : d . D { | × ∈ D}

Proof. Let V be Y -open, so that V = U Y for some U . Consider any x V. Then x U,J and so, since U is -open,∩ ǫ> 0 and d , ...,∈ dJ such that ∈ ∈ D ∃ 1 n ∈D B (x; d , ..., d ) U. ǫ 1 n ⊂

Let dj′ = dj Y Y Y . Then | × ∈D B (x; d′ , ..., d′ )= Y B (x; d , ..., d ) Y U = V. ǫ 1 n ∩ ǫ 1 n ⊂ ∩ Hence x is -interior to V . DY ∴ V is -open. DY ∴ Every -open set is -open. JY DY Conversely, let V be -open. For each x V we can pick a pseudoball DY ∈ B (x; d′ , ..., d′ ) V, ǫ 1 n ⊂ where dj′ = dj Y Y Y . Then U(x)= Bǫ(x; d1, ..., dn) is -open in X, and | × ∈D J U(x) Y = B (x, d′ , ..., d′ ). ∩ ǫ 1 n

Let U = x V U(x). Then U and U Y = V. Hence V Y. ∪ ∈ ∈J ∩ ∈J ∴ Every -open set is -open. DY JY 53 Theorem 16.3. Let (X, ) be a topological space and let Y X. The following statements are equivalent.J ⊂

1. Y is a compact subset of X;

2. (Y, ) is a compact topological space. JY Proof. (1) (2). Suppose Y is a compact subset of X, and so every cover of Y ⇒ by open subsets of X has a ﬁnite subcover. Consider any cover of Y by Y -open sets, say, J = U : α A U { α ∈ } where Uα = Vα Y, Vα . Then Vα : α A is a cover of Y by -open sets. Hence there is a∩ ﬁnite subcover∈ J { ∈ } J

V , ..., V { α1 αn } for some α1, ..., αn A. Then Uα1 , ..., Uαn is a finite subcover of . Hence (Y, ) is compact. ∈ { } U JY (2) (1) Suppose (Y, ) is compact. Consider any cover of Y by -open ⇒ JY U J sets. Then U Y : U { ∩ ∈ U} is a cover of Y by -open sets, hence it has a finite subcover JY U Y, ..., U Y , { α1 ∩ αn ∩ } for some U , ..., U . Clearly U , ..., U is a finite subcover of . α1 αn ∈ U { α1 αn } U ∴ Y is a compact subset of X.

54 17 Product topologies

Let (X , ) and (X , ) be topological spaces. We can deﬁne a topology on 1 J1 2 J2 X1 X2 in a natural way. Consider a subset S of X1 X2 and a point x = (x ,× x ) S. × 1 2 ∈

We say that S is a 1 2-neighbourhood of x if there exists a 1-neighbourhood U of x and a -neighbourhoodJ ×J U of x such that J 1 1 J2 2 2 U U S. 1 × 2 ⊂

We say that a subset S of X1 X2 is 1 2-open if S is a 1 2-neighbourhood of each of its points. × J ×J J ×J

Theorem 17.1. The collection of -open sets is a topology on X X . J1 ×J2 1 × 2

Check the statement.

This topology is called the product topology of 1 and 2, and is denoted by . J J J1 ×J2

[Strictly speaking, this is an abuse of notation, since 1 2 ”ought to” mean (U , U ): U , U .] J ×J { 1 2 1 ∈J1 2 ∈J2}

Example 17.2. Let (X1, 1)=(X2, 2) = R with the natural topology. Then J J 2 1 2 coincides with the natural topology of R , i.e. the topology induced by Jthe×J Euclidean metric on R2.

Let d denote the Euclidean metric on R2. Consider any d-open set U R2 and any x U. Since U is open there exists ǫ > 0 such that B (x; d) U. ⊂Then ∈ ǫ ⊂ (x ǫ , x + ǫ ) is open in R for j =1, 2, and j − √2 j √2 ǫ ǫ ǫ ǫ (x1 , x1 + ) (x2 , x2 + ) Bǫ(x; d) U. − √2 √2 × − √2 √2 ⊂ ⊂ Hence U is a -neighbourhood of each of its points, i.e. U . J1 ×J2 ∈J1 ×J2 Converse – exercise.

55 Theorem 17.3. The product topology comprises the collection of all unions J1×J2 of families of products Uλ Vλ where λ runs over an arbitrary index Λ and Uλ ,V for all λ Λ. × ∈ J1 λ ∈J2 ∈ Proof. Let G be a -open set. For each point λ G, since G is a J1 ×J2 ∈ J1 × 2-neighbourhood of λ, there exist 1- and 2-open sets respectively such that JU V G. Then J J λ × λ ⊂ U V = G λ × λ λ G [∈ and so G is a union of a family of sets U V of the stated type. λ × λ Coversely, suppose G = U V λ × λ λ Λ [∈ for some index set Λ and some families Uλ λ Λ, Vλ λ Λ of 1- and 2- open sets ∈ ∈ respectively. Clearly, for any λ Λ, U { V} is a{ } -neighbourhoodJ J of each ∈ λ × λ J1 ×J2 of its points, i.e. it is 1 2-open. Since 1 2 is a topology, λ Λ Uλ Vλ is J ×J J ×J ∈ × 1 2-open, and hence G is 1 2-open. J ×J J ×J S

Theorem 17.4. If (X, ) and (Y, ) are compact topological spaces then X Y is compact with respect toT the productJ topology . × T ×J

Proof. Let Gλ λ Λ be an open cover of X Y for the product topology . ∈ We shall construct{ } a finite subcover for X ×Y. To begin with hold x X fixedT ×J an consider the points (x, y): y Y . × ∈ { ∈ } For each y Y there is a λ Λ such that (x, y) G and since each G ∈ (x,y) ∈ ∈ λ(x,y) λ is open there are open neighbourhoods Ux(y), Vx(y) of x, y in X,Y respectively such that U (y) V (y) G . x × x ⊂ λ(x,y) Since Vx(y): y Y is an open cover of Y and Y is compact, there is a positive integer{ N(x) and∈ points} y , ..., y Y such that 1 N(x) ∈ V (y ) ... V (y )= Y. x 1 ∪ ∪ x N(x) Let U = U (y ) ... U (y ). x x 1 ∩ ∩ x N(x) Then Ux : x X is an open cover of X. Hence, since X is compact, there is a positive{ integer∈ M}and points x , ..., x X such that 1 M ∈ U ... U = X. x1 ∪ ∪ xM 56 We claim that Uxi Vxi (yj):1 i M, 1 j N(xi) is an open cover of X Y. Since { × ≤ ≤ ≤ ≤ } × U V (y ) G , xi × xi i ⊂ λ(xi,yj ) it will follow that Gλ λ Λ has a finite subcover. { } ∈ M Consider any point (x, y) X Y. Since the (Uxi )i=1 cover X, there exists ∈ × N(xi) i 1, ..., M such that x Uxi . Since, for each i, the sets (Vxi (yj))j=1 cover Y, there∈{ exists j} 1, 2, ..., N(∈x ) such that y V (y ). Thus ∈{ i } ∈ xi j (x, y) U V (y ) ∈ xi × xi j as required.

We have shown that every open subcover of X Y has a ﬁnite subcover. Hence X Y is compact. × ×

57 18 Inﬁnite products

The construction in the previous sections extends fairly easily to arbitary products.

Let Λ be an index set and suppose that, for each λ Λ, we have a topological ∈ space (Xλ, λ). Recall that the Cartesian product of the family Xλ λ Λ is deﬁned ∈ to be the setJ of functions { } x :Λ X → λ λ Λ [∈ such that x(λ) X for all λ Λ. ∈ λ ∈

We often write xλ instead of x(λ), and think of xλ as the ”λth co-ordinate” of x. Thus we can write x =(xλ)λ Λ. ∈

We denote the Cartesian product of the family Xλ λ Λ by λ Λ Xλ. { } ∈ ∈ Q In the case that all the Xλ are equal, say to X, the product λ Λ Xλ is also written XΛ. ∈ Q

We wish to deﬁne a ”product topology” on λ Λ Xλ. For any λ1 Λ and ∈ ∈ U1 λ , let ∈J 1 Q (λ , U )= x X : x U . N 1 1 { ∈ λ λ1 ∈ 1} λ Λ Y∈ For any positive integer n, any λ , ..., λ Λ and 1 n ∈ U for j =1, 2, ..., n, λj ∈Jλj let (λ , ..., λ , U , ..., U )= (λ , U ) ... (λ , U ) N 1 n 1 n N 1 1 ∩ ∩N n n = x X : x U , 1 j n . { ∈ λ λj ∈ j ≤ ≤ } λ Λ Y∈ Consider a subset S of λ Λ Xλ and a point x S. We deﬁne a topology ∈ ∈ Q = J Jλ λ Λ Y∈ on λ Λ Xλ by saying that S is a -neighbourhood of x if there exist a positive ∈ J integer n, points λ1, ..., λn Λ and neighbourhoods Uj λj of xλj for j =1, ..., n suchQ that ∈ ∈J (λ , ..., λ , U , ..., U ) S. N 1 n 1 n ⊂

58 We say that S λ Λ Xλ is -open if S is a -neighbourhood of each of its points. ⊂ ∈ J J We deﬁne toQ be the collection of all -open sets. J J

Theorem 18.1. is a topology on λ Λ Xλ. J ∈ Note that in the case that Λ = 1Q, 2 the topology coincides with the product topology of Section 17.{ } J J1 ×J2

N Example 18.2. Let Λ = N, Xλ = R for all λ N. Here λ Λ Xλ = R the set ∈ ∈ N of inﬁnite sequences x =(xn)n N of real numbers. Any open non-empty set in R ∈ for the product topology is big. Q

Consider, for example, any open neighbourhood U of the zero sequence (0)n N. ∈ For some positive integer m and open neighbourhoods U1, ..., Um of 0 in R we have

U (1, 2, ..., m, U , ..., U ) ⊃N 1 m

= (xn)n N : x1 U1, ..., xm Um . { ∈ ∈ ∈ } Note that only ﬁnitely many co-ordinates of elements of (1, ..., U ) are restricted: N m if (xn)n N then also (x1, x2, ..., xm, xm′ +1, xm′ +2, ...) for any choice of ∈ ∈ N ∈ N x′ , x′ , ... R. m+1 m+2 ∈

N Example 18.3. LetΛ = N,Xλ = [0, 1] for all λ N. Here X = λ Λ Xλ = [0, 1] . ∈ ∈ We can deﬁne a metric d on X by Q

d((xn)n N, (yn)n N) = sup xn yn . ∈ ∈ n | − |

It is not the case that d induces the product topology on [0, 1]N. Consider the sequence x1, x2, x3, ... of elements of [0, 1]N, where

xk = (0, 0, ..., 0, 1, 1, 1, ...) xk tends to the zero sequence as k tends to when [0, 1]N has the product topol- k l ∞ k ogy. However d(x , x ) = 1 whenever k = l, and so the sequence (x )k N does not ∈ converge in the topological space ([0, 1]N ,d).

Remarkably enough the space [0, 1]N is compact with respect to the product topology.

59 Theorem 18.4. (Tychonov’s Theorem) If Λ is a set and (X , ) is a compact λ Jλ topological space for every λ Λ then λ Λ Xλ is compact with respect to the product topology. ∈ ∈ Q

This theorem requires one of the higher axioms of set theory. In fact it is equivalent to the axiom of choice. Thus it is not true in Zermelo-Fraenkel set theory without the axiom of choice. There are models of set theory in which Tychonov’s Theorem is false. However almost all professional mathematicians work within the framework of Zermelo-Fraenkel set theory with the axiom of choice.

60 Part III. The Gelfand representation theorem

The main aim of this part of lectures is to prove the Gelfand representation theorem for a commutative Banach algebra A. To start with we shall deﬁne the character space Aˆ of a commutative Banach algebra A and the Gelfand topology on Aˆ. We shall show that there is a bijective mapping between the character space Aˆ and the set of maximal ideals of A. We shall introduce the radical of A and a notion of semisimplicity of A.

Contents of Part III

Pages 19. The character space 62

20. The Gelfand representation theorem 67

20.1. Singly generated algebras 68

21. Closed ideals 70

22. Quotient algebras 71

23. Maximal ideals and characters 74

24. Semisimplicity and the radical 77

61 19 The character space

Let A be a commutative Banach algebra. Recall that a character of A is a non-zero algebra homomorphism from A to C; the character space Aˆ of A is the set of all characters of A. Lemma 19.1. Let A be a commutative Banach algebra (with or without e). For any character φ of A and any x A, ∈ φ(x) x . | |≤ Proof. Let λ = φ(x) and suppose λ > x . Then x/λ < 1, and so | | x x2 xn y = lim + + ... + n λ λ2 λn →∞ exists; see a proof of Lemma 3.3. We have x x2 xn x x2 xn λy xy = λ lim + + ... + x lim + + ... + − n λ λ2 λn − n λ λ2 λn →∞ →∞ x2 xn x2 xn = lim x + + ... + ... n λ λn 1 − λ − − λn 1 →∞ − − = x. Apply the character φ: λ = φ(x)= φ(λy xy)= λφ(y) φ(x)φ(y) − − = λφ(y) λφ(y)=0. − Therefore, λ = 0. This is a contradiction. Hence φ(x) x . | |≤ Corollary 19.2. Let A be a commutative Banach algebra. The character space Aˆ of A is a subset of the closed unit ball of A∗.

Recall that, for φ A∗, ∈ φ = sup φ(x) . x 1 | | ≤ Definition 19.3. Let A be a commutative Banach algebra. For x A, the Gelfand transform of x is the function ∈ xˆ : Aˆ C : φ φ(x). → → Thus xˆ(φ)= φ(x) for every x A and every character φ of A. ∈ 62 Example 19.4. If A = C(K), then Aˆ = vt : t K . For x C(K), xˆ : Aˆ C satisfies { ∈ } ∈ → xˆ(v )= v (x)= x(t), t K. t t ∈ Thus, if vt is identified with t,x ˆ is the same as x. Example 19.5. If A = ℓ1(Z ), then Aˆ contains the characters φ for α (0, 1) + α ∈△ where ∞ n φα((xn)n∞=0)= xnα . n=0 X Then, if x =(xn)n∞=0, ∞ n xˆ(φα)= φα(x)= xnα . n=0 X Definition 19.6. Let A be a commutative Banach algebra. The Gelfand topology on Aˆ is the coarsest topology for which the functions xˆ : x A are continuous. { ∈ } Recall that if σ, τ are topologies on a set E then σ is coarser than τ is σ τ. ⊂ One also says that τ is finer than σ. Lemma 19.7. If is a collection of C valued functions on a set E then there exists a coarsest topologyF κ for which the− members of are continuous. A base F for κ consists of the sets V (f , ..., f ; U , ..., U )= t E : f (t) U , 1 j n 1 n 1 n { ∈ j ∈ j ≤ ≤ } where n N, f , ..., f and U , ..., U are open in C. ∈ 1 n ∈F 1 n Proof. One can readily check that the sets V (f1, ..., fn; U1, ..., Un) do comprise the base for a topology κ on E. For any f and open U C we have ∈F ⊂ 1 f − (U)= V (f; U), a basic κ-open set. Hence the members of are continuous with respect to k. Let τ be a topology on E such that allF members of are continuous with respect to τ. Consider any f , f , ..., f and open U , ...,F U in C. Then 1 2 n ∈F 1 n 1 f − (U ) τ, j j ∈ and hence

1 1 V (f , ..., f ; , U , ..., U )= f − (U ) ... f − (U ) τ. 1 n 1 n 1 1 ∩ ∩ n n ∈ Thus κ τ. Therefore κ is the coarsest topology for which the members of are continuous.⊂ F

63 Thus the deﬁnition of the Gelfand topology in Deﬁnition 19.6 makes sense, and we have:

Theorem 19.8. Let A be a commutative Banach algebra. A base for the Gelfand topology on Aˆ is given by the sets

V (x , ..., x ; , U , ..., U )= φ Aˆ :x ˆ (φ) U , 1 j n 1 n 1 n { ∈ j ∈ j ≤ ≤ } = φ Aˆ : φ(x ) U , 1 j n { ∈ j ∈ j ≤ ≤ } where n N, x , ..., x A, U , ..., U are open in C. ∈ 1 n ∈ 1 n Note that, by Lemma 19.1, any character φ of A satisﬁes

φ(x) (0, x )= z C : z x ∈△ { ∈ | |≤ } for x A. Hence ∈ φ (0, x ). ∈ △ x A Y∈ Corollary 19.9. Let A be a commutative Banach algebra. The Gelfand topology ˆ ˆ of A coincides with the relative topology A enjoys as a subset of x A (0, x ) with the product topology. ∈ △ Q

Proof. The sets V (x1, ..., xn; U1, ..., Un) in Theorem 19.8 constitute a base for both topologies.

Theorem 19.10. If the commutative Banach algebra A has an identity then Aˆ is a compact Hausdorﬀ space with respect to the Gelfand topology.

Proof. Let A have identity e. Let H be the set of homomorphisms from A to C. Note that H = Aˆ 0 . ∪{ } Let Y = x A (0, x ). By Tychonov’s Theorem 18.4, Y is compact and Hausdorﬀ in the∈ product△ topology. Claim: H Qis a closed subset of Y . Let x A and ∈ v : Y C : φ φ(x). x → → For U open in C,

1 v− (U)= φ (0, y ): φ(x) U x { ∈ △ ∈ } y A Y∈ is a basic open set in Y, and so vx is continuous.

64 Hence, for any x, y A, ∈ v v v : Y C xy − x y → is continuous. Therefore

1 (v v v )− ( 0 ) xy − x y { } x,y A \∈ is closed in Y . That is, the set of multiplicative functions in Y is closed in Y . Similarly for the other algebraic operations. Thus H is a closed subset of the compact Hausdorff space Y . By Theorem 15.7, H is a compact Hausdorff space. For φ H we have ∈ φ Aˆ φ(e)=1, ∈ ⇐⇒ so that 1 Aˆ = H v− ( 1 ). ∩ e { } Hence Aˆ is a closed subspace of H, and so Aˆ is a compact Hausdorff space.

Theorem 19.11. For any compact Hausdorﬀ space K, the character space C(K)Λ of C(K) in the Gelfand topology is homeomorphic to K.

Proof. By Theorem 9.2, the map

v : K C(K)Λ : t v → → t is surjective. A theorem in topology (Urysohn’s Lemma) asserts that if t = t in 1 2 K then there exists f C(K) such that f(t1) = f(t2), and hence vt1 = vt2 . Thus v is bijective. ∈ For any x C(K), we have, for t K, ∈ ∈ 1 xˆ(v )= x(t)= x v− (v ), t ◦ t so that 1 xˆ = x v− . ◦ Let τ be the ”topology of K transferred to C(K)Λ”: more precisely,

τ = v(U): U open in K . { } Claim: for any x C(K), ∈ xˆ :(C(K)Λ, τ) C →

65 is continuous. For any open set W C we have ⊂ 1 1 1 1 xˆ− (W )=(x v− )− (W )= v(x− (W )) τ ◦ ∈ 1 since x− (W ) is open in K. Hencex ˆ is continuous with respect to τ. Since the Gelfand topology is the coarsest topology for which all thex ˆ are continuous, τ is ﬁner than the Gelfand topology. That is, for each Gelfand-open set V in C(K)Λ, 1 Λ v− (V ) is open in K. Hence v : K C(K) is continuous. Every bijective continuous map→ from a compact space to a Hausdorﬀ space is a homeomorphism, Hence K is homeomorphic to C(K)Λ. This theorem shows that we can ”recover” K from the structure of the algebra C(K).

66 20 The Gelfand representation theorem

To what extent is a general commutative Banach algebra like C(K) for some Hausdorff space K? Theorem 20.1. Let A be a commutative Banach algebra with identity. The Gelfand transformation Γ: A C(Aˆ): x xˆ → → is an algebra homomorphism from A to the algebra of continuous C-valued functions on the compact Hausdorff space Aˆ in its Gelfand topology, with pointwise operations. Moreover Γ 1. ≤ Proof. By Theorem 19.10, Aˆ is a compact Hausdorf space. For x A,x ˆ is ∈ continuous on Aˆ by Definition 19.6, that is,x ˆ C(Aˆ). Thus Γ does map A into C(Aˆ). ∈ Note that Γ is linear. Consider x, y A and λ C. For any φ Aˆ, we have ∈ ∈ ∈ (x + y)(φ)= φ(x + y)= φ(x)+ φ(y)=ˆx(φ)+ˆy(φ).

∴ (x + y)=ˆx +ˆy. ∴ Γ(x + y)=(Γx)+(Γy) for all x, y A. Therefore Γ is additive. ∈ λxφ = φ(λx)= λxˆ(φ)

c∴ Γ(λx)= λΓ(x) for all x A and λ C. Thus Γ is multiplicative. ∈ ∈ For φ A,ˆ ∈ (xy)(φ)= φ(xy)= φ(x)φ(y)=ˆx(φ)ˆy(φ)=(ˆxyˆ)(φ).

d ∴ Γ(xy)=(Γx)(Γy) for all x, y A. Hence Γ is a homomorphism of algebras. For x ∈A, we have ∈ Γx C(Aˆ) = sup Γx(φ) φ Aˆ | | ∈ = sup xˆ(φ) = sup φ(x) φ Aˆ | | φ Aˆ | | ∈ ∈ x (by Lemma 19.1). ≤ A Hence Γ 1. ≤ 67 As we have seen in Theorem 19.11, if A = C(K) for some compact Hausdorff space K then Aˆ can be identified with K and Γ is the identity mapping A C(Aˆ). More commonly Range Γ is a proper subalgebra of C(Aˆ). → Example 20.2. Let A = C1[0, 1], the Banach algebra of continuously differentiable C-valued functions on [0, 1] with pointwise operations and norm

f C1 = sup f(t) + sup f ′(t) . 0 t 1 | | 0 t 1 | | ≤ ≤ ≤ ≤ Then Aˆ consists of evaluation functionals

v : A C : f f(t) t → → for 0 t 1. The character space Aˆ in its Gelfand topology can be identiﬁed with [0, 1]≤ in its≤ standard topology, and Γ : C1[0, 1] C[0, 1] is the natural injection mapping. → We shall shortly give an example in which Γ is not an obvious injection.

20.1 Singly generated algebras. Deﬁnition 20.3. Let A be a Banach algebra with identity e. We say that A is generated by an element x A if the smallest closed subalgebra of A containing e and x is A; equivalently, if∈ the closed linear span of e, x, x2, ... in A is A. { } Theorem 20.4. Let A be a commutative Banach algebra with identity e and suppose that A is generated by x. The character space Aˆ can be identiﬁed with the spectrum σ(x) of x, and the Gelfand topology of Aˆ agrees with the natural topology of σ(x) C. More precisely, ⊂ xˆ : Aˆ σ(x) C : φ xˆ(φ)= φ(x) → ⊂ → is a homeomorphism. Proof. Proof to follow on Page 76.

1 Example 20.5. Consider ℓ (Z+) with convolution multiplication (Example 6.1). What is the Gelfand representation of A? Note that A is singly generated. Let en = (0, ..., 0, 1, 0, ...) (1 in the nth place for n Z+.) The element e0 is an identity for A and e1 is a generator. In fact e e ∈ ... e = e , and 1 ∗ 1 ∗ ∗ 1 n (x , x , ..., x , 0, 0, ...)= x e + x e + x e e + ... + x e ... e . 0 1 n 0 0 1 1 2 1 ∗ 1 n 1 ∗ ∗ 1 68 Hence the smallest closed subalgebra of A containing e0 and e1 is A. It follows that Aˆ is homeomorphic to σ(e ). Since e =1, we have σ(e ) (0, 1). 1 1 1 ⊂△

Claim: each φ Aˆ has a form φα for some α (0, 1). Consider any ∈ ∈ △ 1 φ Aˆ and suppose φ(e1) = α (0, 1). For x = (x0, x1, ..., xn, 0, 0, ...) ℓ (Z+), we∈ have ∈ △ ∈ x = x e + x e + x e e + ... + x e ... e 0 0 1 1 2 1 ∗ 1 n 1 ∗ ∗ 1 and so

2 n φ(x)= x0φ(e0)+ x1φ(e1)+ x2φ(e1) + ... + xnφ(e1) n j = xjα j=0 X = φα(x) in the notation of Example 7.2(5).

Since x A : φ(x) = φ (x) is a closed subalgebra of A that contains e and e { ∈ α } 0 1 it equals A; that is, φ = φα. Hence

eˆ : Aˆ σ(e )= (0, 1) : φ φ (e )= α 1 → 1 △ α → α 1 is a homeomorphism, and if we identity Aˆ with (0, 1) via φ α we have the Gelfand representation △ →

Γ: ℓ1(Z ) C( (0, 1)) : x xˆ + → △ → where ∞ n xˆ(α)= φα(x)= xnα . n=0 X 1 Notice that if (x )∞ ℓ (Z ) then n n=0 ∈ +

∞ n xˆ(z)= xnz n=0 X is an absolutely convergent Taylor series in (0, 1). △ 1 Thus ”Γ maps ℓ (Z+) to the algebra of analytic functions in D with absolutely convergent Taylor series in (0, 1).” △

69 21 Closed ideals

Let A be a Banach algebra with identity e. Ideals of A are not necessarily closed. For example, in ℓ∞ the linear subspace cF of ﬁnitely non-zero sequences is an ideal, but its closure is c0. Lemma 21.1. Let A be a Banach algebra with identity e. (i) The closure of an ideal in A is an ideal. (ii) The closure of a proper ideal is a proper ideal.

Proof. (i). Let I be an ideal, we denote the closure of I by Cl I. Consider x, y Cl I. Suppose x + y / Cl I. There is a neighbourhood U of x + y which does ∈ ∈ not meet I. Since addition A A A × → is continuous at (x, y) there are neighbourhood V, W of x, y respectively such that V + W U. Since x, y Cl I there exist points ξ V I, η W I. Then ξ + η V⊂ + W U and∈ξ + η I, contradicting the∈ fact∩ that U∈ I =∩ . Hence ∈ ⊂ ∈ ∩ ∅ x + y Cl I. Likewise,∈ for a A and x Cl I, xa ClI and ax ClI, and λx ClI for λ C. Hence ClI ∈is an ideal.∈ ∈ ∈ ∈ ∈ (ii). Let I be a proper ideal of A. By Theorem 7.12, I does not contain any regular element of A. By Lemma 3.3, B1(e) is a neighbourhood of e consisting of regular elements, hence is disjoint from I. Thus e / Cl I, and so ClI is a proper ideal. ∈

Theorem 21.2. Let A be a Banach algebra with identity e. Maximal ideals of A are closed.

Proof. Let I be a maximal ideal of A. By deﬁnition I is proper, and so, by Lemma 21.1, ClI is a proper ideal of A. We have

I ClI A, ⊂ and so, by maximality of I, I = Cl I. Thus I is closed.

70 22 Quotient algebras

Recall that the kernel of an algebra homomorphism is an ideal, see Remark 7.9. We shall show the converse: every ideal is the kernel of a homomorphism. Given an ideal I in an algebra A we shall construct an algebra A/I (”A mod I”) and a homomorphism φ : A A/I such that ker φ = I. → Deﬁnition 22.1. Let A be an algebra and let I be an ideal in A. For x A the ∈ coset of x mod I is the set x + I, that is, x + y : y I . The quotient set A/I is the set of cosets{ in A (mod∈ }I):

A/I = x + I : x A . { ∈ } Note: if we write x y (mod I) to mean x y I, then (mod I) is an equivalence relation, and≡A/I consists of the corresponding− ∈ equivalence≡ classes.

Example 22.2. Let A = C[0, 1] and let 1 I = f A : f(t)=0 for t 0, . { ∈ ∈ 2 } I is an ideal of A. For any g A, g + I consists of all f A such that f agrees 1 ∈ ∈ 1 with g on 0, 2 . A/I can be naturally identiﬁed with C 0, 2 . Theorem 22.3. Let A be an algebra and let I be an ideal in A. The quotient linear space A/I is an algebra with respect to the operations deﬁned for x, y A and scalar λ C by ∈ ∈ (x + I)+(y + I)=(x + y)+ I, (x + I)(y + I)=(xy)+ I, λ(x + I)=(λx)+ I.

Furthermore the map k : A A/I : x x + I → → is a homomorphism of algebras.

Proof. The above operations are well deﬁned, for if x+I = x′ +I and y+I = y′ +I then there are i , i I such that 1 2 ∈ x x′ = i , y y′ = i , − 1 − 2 and hence x + y (x′ + y′)= i + i I, so that − 1 2 ∈ (x + y)+ I =(x′ + y′)+ I.

71 Similarly

x′y′ xy =(x i )(y i ) xy = i y xi + i i I, − − 1 − 2 − − 1 − 2 1 2 ∈ so that x′y′ + I = xy + I, and λx λx′ = λi I, − 1 ∈ so that λx + I = λx′ + I. The algebra axioms for A/I follow easily from those for A. The fact that k is a homomorphism is immediate from the deﬁnition.

Deﬁnition 22.4. k : A A/I is called the quotient map or the canonical homomorphism. →

Corollary 22.5. Let A be an algebra. The ideals of A coincide with the kernels of homomorphisms of A.

Proof. If φ : A B is a homomorphism then ker φ is an ideal. If I is an ideal then → k : A A/I → is a homomorphism of algebras and

x ker k k(x) = 0 in A/I ∈ ⇐⇒ x + I =0+ I ⇐⇒ x I. ⇐⇒ ∈ Thus I = ker k, and so I is the kernel of a homomorphism. Now consider the case of normed algebras.

Deﬁnition 22.6. Let E be a normed space and let F be a closed subspace of E. For x E the coset of x mod F is the set x + F. The quotient set E/F is the set of cosets∈ of E mod F . Addition and scalar multiplication are deﬁned in E/F by

(x + F )+(y + F )=(x + y)+ F, λ(x + F )=(λx)+ F. (22.1)

72 Theorem 22.7. Let E be a normed space and let F be a closed subspace of E. The quotient set E/F is a linear space under operations (22.1). The pair (E/F, . E/F ) is a norm space, where

x + F E/F = inf x + y E. y F ∈ If E is a Banach space then so is E/F.

. is called the quotient norm. E/F Corollary 22.8. If I is a closed ideal in a Banach algebra A then A/I is a Banach algebra under the quotient norm. If A has an identity, so does A/I.

Proof. By Theorems 22.5 and 22.7, A/I is an algebra and a Banach space. For x, y A, ∈ (x + I)(y + I) = xy + I A/I A/I = inf xy + z A z I ∈ inf (x + z1)(y + z2) A z1,z2 I ≤ ∈ inf x + z1 A y + z2 A z1,z2 I ≤ ∈ = x + I y + I . A/I A/I Thus . is submultiplicative. Hence A/I is a Banach algebra. A/I If e is an identity in A then e + I is one in A/I.

Example 22.9. Let A = C[0, 1] and let

1 I = f A : f(t)=0 for t 0, . { ∈ ∈ 2 } I is an ideal of A as in Example 22.2. Deﬁne

1 φ : A/I C 0, : f + I f [0, 1 ]. → 2 → | 2 Then φ : A/I C 0, 1 is an isometric isomorphism. → 2

73 23 Maximal ideals and characters

Recall Theorem 7.15 that the kernel of a character of a commutative Banach algebra is a maximal ideal.

Theorem 23.1. Every maximal ideal of a commutative Banach algebra with identity is the kernel of a character.

Proof. Let A be commutative Banach algebra with identity e and let I be a maximal ideal of A. By Theorem 21.2, I is a closed ideal. By Corollary 22.8, A/I is a commutative Banach algebra with identity. The zero element of A/I is 0 + I, which we can write I . Claim: I is a maximal ideal of A/I. Let { } { } k : A A/I : x x + I → → be the canonical homomorphism. Consider any proper ideal of A/I. Then J 1 k− ( )= x A : x + I J { ∈ ∈J} 1 is a proper ideal of A containing I. Since I is maximal we have k− (J)= I, that is, x + I x I. ∈J ⇐⇒ ∈ Therefore, = I . Hence I is a maximal ideal, as claimed. J { } { } Since A/I is a commutative Banach algebra with identity in which the only ideals are I and A/I, it follows from Examples Sheet 5, Q7, that A/I is a ﬁeld and so, by{ the} Gelfand-Mazur Theorem 5.5, that A/I is isomorphic (as an algebra) to C. Hence k : A A/I C is a character (note that k(e) is the identity of A/I, → ≈ so k = 0), and I = ker k is the kernel of a character. Corollary 23.2. There is a bijective mapping φ ker φ between the character space and the set of maximal ideals of a commutative→ Banach algebra with identity.

We shall denote by MA the set of maximal ideals of a commutative Banach algebra A. Proof. By Theorem 7.15, Aˆ M : φ ker φ → A → is a well deﬁned mapping, and, by Theorem 23.1, it is surjective. It is also injective. Suppose φ, ψ Aˆ and ker φ = ker ψ. We have ∈ φ(e)= ψ(e)=1

74 where e is the identity of A. For any x A, x φ(x)e ker φ = ker ψ, and so ∈ − ∈ 0= ψ(x φ(x)e)= ψ(x) φ(x)ψ(e)= ψ(x) φ(x). − − − Hence φ = ψ. Thus Aˆ M : φ ker φ is a bijection. → A → Remark 23.3. In view of Corollary 23.2 we can identify Aˆ with MA. For any x A, we can regard the Gelfand transformx ˆ as a function ∈ xˆ : M C, A → given by xˆ(M)= φ(x), where M = ker φ M . ∈ A The Gelfand topology on MA is the coarsest topology for which the functions x,ˆ x A, are continuous. ∈ Deﬁnition 23.4. Let A be commutative Banach algebra with identity e. The set of maximal ideals MA, with the Gelfand topology, is called the maximal ideal space of A. Corollary 23.5. In any commutative Banach algebra A with identity, for any x A, ∈ σ(x)= φ(x): φ Aˆ = Rangex. ˆ { ∈ } Proof. By Theorem 7.5, φ(x): φ Aˆ σ(x). { ∈ } ⊂ Consider λ σ(x). The ideal (λe x)A is proper, hence, by Theorem 7.17, (λe x)A is contained∈ in a maximal ideal−M. By Theorem 23.1, there is a character − φ of A such that M = ker φ. Then λe x ker φ, that is − ∈ φ(λe x)=0. − Hence φ(x)= φ(λe)= λ. Thus σ(x) φ(x): φ Aˆ . ⊂{ ∈ } Hence σ(x)= φ(x): φ Aˆ { ∈ } = xˆ(φ): φ Aˆ { ∈ } = Rangex. ˆ

75 Proof of Theorem 20.4: if a commutative Banach algebra with identity e is generated by an element x then

xˆ : Aˆ σ(x) → is a homeomorphism. Proof. By Corollary 23.5,x ˆ : Aˆ σ(x) is a surjective mapping. It is also injective, for suppose φ, ψ Aˆ andx ˆ(φ)=ˆ→x(ψ). Then ∈ a A : φ(a)= ψ(a) { ∈ } is a closed subalgebra of A containing e and x, hence it is all of A. That is, φ = ψ. It follows thatx ˆ : Aˆ σ(x) is a bijective mapping. The functionx ˆ is continuous, by deﬁnition of the Gelfand→ topology. The character space Aˆ is compact and σ(x) is Hausdorﬀ, hencex ˆ is a homeomorphism.

76 24 Semisimplicity and the radical

Return to the Question of Section 20: to what extent does a general commutative Banach algebra resemble C(K)? In the case that A has an identity we answered the question by constructing the Gelfand transformation

Γ: A C(Aˆ), → so that A can be homomorphically mapped into C(Aˆ). If Γ is injective, this means that A can be regarded as a subalgebra of C(Aˆ). However, Γ need not be injective. Examples 24.1. 1. Let E be a Banach space endowed with zero multiplication: xy = 0 x, y E. Then E is a commutative Banach algebra without ∀ ∈ identity. Note that Eˆ is empty. For suppose φ is a character of E. For x E, we have ∈ 2 φ(x )= φ(xx)= φ(0) = 0, and so φ = 0 - a contradiction. Thus Γ is not deﬁned for E. 2. Let A be the algebra obtained by ”adjoining an identity” to E. That is, A = C E, with operations ⊕ (λ, x)+(,y)=(λ + , x + y),

(λ, x)(,y)=(λ, x + λy), c(λ, x)=(cλ, cx), for λ,,c C and x, y E. ∈ ∈ A is a Banach algebra with identity under the norm

(λ, x) = λ + x . | | E Its identity is e = (1, 0).

We can deﬁne a character φ0 on A by

φ0(λ, x)= λ. If ψ is any character on A then ψ E is a homomorphism E C, hence is 0. Thus | → ψ(λ, x)= ψ(λe + (0, x)) = λψ(e)= λ, and so ψ = φ . Thus Aˆ = φ . Here C(Aˆ) is one-dimensional, and 0 { 0} Γ: A C(Aˆ) → is far from injective. In fact ker Γ = E.

77 3. A more natural example than (2): A = C L1(0, 1), with convolution multiplication also has a unique character ⊕

(λ, x) λ. → When ker Γ is large the Gelfand transformation tells us little about the structure of an algebra A. However, for many natural algebras, ker Γ = 0 . { } Deﬁnition 24.2. Let A be a commutative algebra with identity. The Jacobian radical RadA of A is deﬁned to be the intersection of all the maximal ideals of A. It is an ideal of A.

Note that if A is a commutative Banach algebra with identity then RadA is a closed ideal of A, since maximal ideals are closed.

Theorem 24.3. For any commutative Banach algebra A with identity the kernel of the Gelfand transformation Γ of A is RadA.

Proof. Consider x A. ∈ x ker Γ Γx =0 ∈ ⇐⇒ xˆ =0 ⇐⇒ xˆ(φ)=0 φ Aˆ ⇐⇒ ∀ ∈ φ(x)=0 φ Aˆ ⇐⇒ ∀ ∈ x ker φ φ Aˆ ⇐⇒ ∈ ∀ ∈ x M for all M M ⇐⇒ ∈ ∈ A x M ⇐⇒ ∈ M MA \∈ x RadA. ⇐⇒ ∈ Hence ker Γ = Rad A

Corollary 24.4. The Gelfand transformation of a commutative Banach algebra A with identity is an injective homomorphism if and only if Rad A = 0 . { } Deﬁnition 24.5. A commutative Banach algebra with identity is semisimple if its Jacobian radical is 0 . { } Thus A is semisimple Rad A = 0 Γ is injective. ⇐⇒ { } ⇐⇒ Intuitively, A can be thought of as an algebra of continuous functions with pointwise operations if and only if A is semisimple.

78 Examples 24.6. 1. Any algebra which is deﬁned as an algebra of functions with pointwise operations is semisimple. Indeed, any function in the radical lies in the kernel of all point evaluations, hence is identically zero.

Thus ℓ∞ is semisimple.

1 2. A = ℓ (Z+) is semisimple. Recall Example 20.5,

Aˆ = (0, 1) = z C : z 1 . △ { ∈ | | ≤ } 1 Consider x Rad ℓ (Z ), x =(x )∞ . We have, for all z (0, 1), ∈ + n n=0 ∈△

∞ n φz(x)=ˆx(z)= xnz =0. n=0 X 1 Notice that if (x )∞ ℓ (Z ) then n n=0 ∈ +

∞ n xˆ(z)= xnz n=0 X

is an absolutely convergent Taylor series in (0, 1). It follows that all xn = 0, for example, from the relation △

1 xˆ(z) xn = n+1 dz. 2π z =1 z Z| |